Energy Problems

Question A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg mass sitting still. Given that the collision is perfectly elastic, determine the final velocities of each of the masses.

Solutions Question A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg mass sitting still. Given that the collision is perfectly elastic, determine the final velocities of each of the masses. Conservation of Momentum Conservation of Energy Oh No, we have 2 unknowns. If only there was another equation.

Question A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg mass sitting still. Given that the collision is perfectly elastic, determine the final velocities of each of the masses.

Example 0 A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. a) How much work is done by gravity? b) How much work is done by the normal force? c) How much work is done by friction? d) What is the total work done?

Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. a) How much work is done by gravity? Recall Work = force x distance with the force being parallel to the distance, that is, the component down the ramp Fg

Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. b) How much work is done by the normal force? Since the normal force is perpendicular to the displacement, NO WORK IS DONE.

Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. c) How much work is done by friction? Recall Work = force x distance.

Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. d) What is the total work done? Total Work = sum of all works

Example 2 A 2.00kg block is pushed against a spring with negligible mass and a force constant k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a frictionless horizontal surface and then up a frictionless incline with a slope of 37.00. a) What is the speed of the block as it slides along the horizontal surface after having left the spring? b) How far does the block travel up the incline before starting to slide back down?

Solution to Example 2 Let’s find the Spring Potential Energy first. A 2.00kg block is pushed against a spring with negligible mass and a force constant k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a frictionless horizontal surface and then up a frictionless incline with a slope of 37.00. a) What is the speed of the block as it slides along the horizontal surface after having left the spring? b) How far does the block travel up the incline before starting to slide back down? Let’s find the Spring Potential Energy first. When released the Spring Potential Energy is transferred into kinetic energy of the Block, where we can solve for v

Solution to Example 1 A 2.00kg block is pushed against a spring with negligible mass and a force constant k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a frictionless horizontal surface and then up a frictionless incline with a slope of 37.00. a) What is the speed of the block as it slides along the horizontal surface after having left the spring? b) How far does the block travel up the incline before starting to slide back down? We can use conservation of Spring Potential and Gravitational Potential Energy Using Trig, we will translate the vertical distance to a ramp distance

Question 2 The planet Saturn is moving in the negative x-direction at its orbital speed (with respect to the Sun) of 9.6 km/s. The mass of Saturn is 5.69x1026 kg. A 2150 kg spacecraft approaches Saturn, moving initially in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) causes the spacecraft to swing around it and head of into the opposite direction. Determine the speed of the spacecraft after it is far enough away to be free of Saturn’s gravitational pull.

Solution to Question 2 The planet Saturn is moving in the negative x-direction at its orbital speed (with respect to the Sun) of 9.6 km/s. The mass of Saturn is 5.69x1026 kg. A 2150 kg spacecraft approaches Saturn, moving initially in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) causes the spacecraft to swing around it and head of into the opposite direction. Determine the speed of the spacecraft after it is far enough away to be free of Saturn’s gravitational pull. Here the “collision” is not an impact but rather a gravitational interaction. So we can treat it as a one-dimensional elastic collision. Therefore the relative velocities before and after the collision have the same magnitude but opposite sign. The reason the spacecraft acquires so much additional speed is that Saturn is moving in its orbit. It does slow down, but its mass is so much greater than the spacecraft, it is not noticeable.

Question 4 The figure below shows a ballistic pendulum, the system for measuring the speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of M=3.0kg , suspended like a pendulum, and makes a completely inelastic collision with it. After the impact of the bullet, the block swings up to a maximum height 2.00 cm. Determine the initial velocity of the bullet? Stage 1 (conservation of Momentum) Stage 2 (conservation of Energy) Therefore:

Question 6 A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional force to the elevator. a) Determine the force constant of the spring.

The elevator’s initial kinetic energy is: Solution to Question 6 A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional force to the elevator. a) Determine the force constant of the spring. The elevator’s initial kinetic energy is: If we use Point 1 to be the origin, then we have Us=0 and Ugrav=0, and so U1=0 At Point 2, there is both gravitational and elastic potential energy, but no kinetic energy Putting the conservation of energy to work: K1+U1+Wother=K2+U2

Solution to Question 6 A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional force to the elevator. a) Determine the force constant of the spring. If we use Point 1 to be the origin, then we have Us=0 and Ugrav=0, and so U1=0

Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, i) as it passes through the equilibrium point? ii) when it has moved 5.0 cm from its release point? b) Find the time after release that the block first passed through the position x = 3.0 cm (on the opposite side of its equilibrium point). 0.16m

Example 10 Solution A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, i) as it passes through the equilibrium point? We can solve this in two different ways 1) Using Energies 0.16m

2) Using Velocity Function Example 10 Solution A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, i) as it passes through the equilibrium point? We can solve this in two different ways 2) Using Velocity Function 0.16m ¼ of T We need only the time when the mass is at the equilibrium point (ie x=0). We can get that by setting the position function = 0 and solving for time or by using one quarter of the period.

Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, ii) when it has moved 5.0 cm from its release point? We can solve this in two different ways 1) Using Energies 5cm 16 cm x is the position that the spring is stretched from the equilibrium at that at that velocity. This distance is 16cm – 5 cm= 11 cm.

2) Using Velocity Function Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, ii) when it has moved 5.0 cm from its release point? We can solve this in two different ways 2) Using Velocity Function 5cm 16 cm Setting the position functions to 0.11 m and solving for time

Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. b) Find the time after release that the block first passed through the position x = 3.0 cm (on the opposite side of its equilibrium point). This is a job for the position function. 3 cm 16 cm

Example 4 A springs (k1=100 N/m and k2=300 N/m) are connected in series. Determine the amount of force required to pull them 6cm apart. Determine the effective k value for springs in series. Then use the normal formulas Since xeq=x1+x2, then: since Forces are all the same:

Example 4 A springs (k1=100 N/m and k2=300 N/m) are connected in series. Determine the amount of force required to pull them 6cm apart. Now Apply F=kx. Therefore:

Example 4 A springs (k1=10 N/m and k2=30 N/m) are connected in parallel. Determine the Period when they are to pulled 6cm apart when a 10kg mass is attached. Determine the effective k value for springs in parallel. Then use the normal formulas Right Force = sum of Left Forces Plug this into:

Example 4 A 10kg mass is attached between two springs (k1=10 N/m and k2=30 N/m) . The mass is pulled to the right 6cm apart. Determine the Period of the oscillation when the mass is released. Determine the effective k value for springs in parallel. Then use the normal formulas pulling pushing When pulled to the right, both springs will have a force to the left (one pulling, the other pushing) Plug this into:

Example 5 A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m. The collision is elastic. a. When the pendulum reaches the vertical position, calculate the speed of the bob just before it strikes the box. b. Calculate the speed of the bob and the box just after they collide elastically. c. Determine the impulse acting on the box during the collision. d. Determine how far away from the bottom edge of the table, Δx, the box will the box strike the floor.

Example 5 A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m. a. When the pendulum reaches the vertical position, calculate the speed of the bob just before it strikes the box.

Example 5 A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m. b. Calculate the speed of the bob and the box just after they collide elastically.

Example 5 A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m. c. Determine the impulse acting on the box during the collision.

Example 5 A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m. d. Determine how far away from the bottom edge of the table, Δx, the box will the box strike the floor.

Example 5 At the location where the box would have struck the floor, now a small cart of mass M = 5.0 kg and negligible height is placed. The box lands in the cart and sticks to the cart in a perfectly inelastic collision. Calculate the horizontal velocity of the cart just after the box lands in it.