During a museum refit, a wheeled display cabinet is moved along a straight guide rail. A constant force of 16 N acts along the rail. The position sensor reads 125 cm before the movement and 175 cm afterwards.
A trainee records the work done as
E=16÷(175−125)=0.32 JE=16\div(175-125)=0.32\ \mathrm{J}E=16÷(175−125)=0.32 J
By what factor must the trainee's value be multiplied to obtain the correct work done?
100100100
125\dfrac{1}{25}251
252525
250025002500
115 exam-style questions on Edexcel GCSE Physics 8.1 Energy transfers and work, covering 8.1.1 Energy stores and transfer diagrams, 8.1.2 Ways of changing a system's energy, and 8.1.3 Work done by forces. Each one has a worked solution and a mark scheme showing where the marks go.