Gravitational field strength = 9.81 m/s2 on Earth Forces A force is a push or a pull. Unit: Newton (N). Two types of force: Contact: force experienced by an object when it is in contact with another object, e.g. drag. Non-contact: force experienced by an object even when it is not in contact with another, e.g. weight, magnetic. Forces are vectors: they have size (or magnitude) and direction. Other vectors include displacement, velocity and acceleration. The opposite of a vector is a scalar, which only has size, e.g. speed, distance and time. Weight (W) is the force on an object due to gravity: Question: Find W for a 700g mass on Earth. W = mg = (700/1000) x 9.81 = 6.9N W = mg Weight in Newtons, N Gravitational field strength = 9.81 m/s2 on Earth Mass in kilograms, kg 4.5.1.1 – 4.5.1.3

Resultant Forces & Work Done A number of forces acting on an object may be replaced by a single force that has the same effect as all the original forces acting together. This single force is called the resultant force. When a force causes an object to move through a distance work is done (W) on the object. So a force does work on an object when the force causes a displacement of the object: W = Fd Work done in Joules, J Distance perpendicular to force in metres, m Force in Newtons, N 4.5.1.4 – 4.5.2

Springs A stretched string pulls on the object holding each end. This is known as the tension in the spring, and is equal and opposite to the force needed to stretch it. Hooke’s law states that the force (F) needed to stretch a spring is directly proportional to the extension of the spring from its natural length (e): Springs will obey Hooke’s law up to a point known as their limit of proportionality. During this time, they undergo elastic deformation (because if the force is removed, they’d regain their original shape). Beyond its elastic limit, the spring is permanently stretched and undergoes plastic deformation. Question: Find the force needed to stretch a spring from its original length of 6.5cm to a length of 13.9cm. The spring constant for this spring is 40N/m. Give your answer to 1 sig. figure. F = kΔx = 40 x (13.9x10-2 – 6.5x10-2) = 2.96N = 3N F = ke Force in Newtons, N Spring constant in N/m Extension in metres, m 4.5.3

Elastic Potential Energy The elastic potential energy (energy stored in a spring or any elastic object) is given by: (Note that this equation will be on the formulae sheet in the exam.) Question: Calculate the spring constant for a spring that, when stretched from an original length of 5cm to an extended length of 14cm, stores 4,500J of elastic potential energy. E = ½ke2 k = 2E ÷ e2 = (2 x 4500) ÷ ((14-5)/100)2 = 1,111,111 N/m E = ½ke2 Force in Newtons, N Spring constant in N/m Extension in metres, m 4.5.3