Question

The only force acting on a 2.8 kg canister that is moving in an xy plane has a magnitude of 5.8 N. The canister initially has a velocity of 4.4 m/s in the positive x direction, and some time later has a velocity of 6.7 m/s in the positive y direction. How much work is done on the canister by the 5.8 N force during this time?

Answer 1

Explanations & Calculations

• The magnitude of the speed has changed over that time.
• Hence there is a change in kinetic energy and it should be a gain in kinetic energy as the final speed is larger than the initial one.
• That difference is what the force has done on the object assuming that no other friction forces apply.
• Then,

$\qquad\qquad \begin{aligned} \small Work&=\small \Delta k.e\\ &=\small \frac{1}{2}m(v_f^2-v_i^2)\\ &=\small 0.5\times2.8(6.7^2-4.4^2)\\ &=\small 35.7\,J \end{aligned}$