Question 1

Paper 2Medium6 marks

A trolley of mass 24 kg24\ \mathrm{kg} is pulled horizontally along a level track by a rope. The rope exerts a forward force of 48 N48\ \mathrm{N} while the track exerts a frictional force of 12 N12\ \mathrm{N} backwards. These forces remain constant for 4.0 s4.0\ \mathrm{s} At the start of this interval, the trolley is moving forwards at 1.5 m s−11.5\ \mathrm{m\,s^{-1}} Ignore any other horizontal forces.

The rope force is then reduced, and the trolley continues at constant velocity. The frictional force remains 12 N12\ \mathrm{N}

(a)

Calculate the trolley’s acceleration during the 4.0 s4.0\ \mathrm{s} interval.

[2]
(b)

Determine the trolley’s speed at the end of the 4.0 s4.0\ \mathrm{s} interval.

[1]
(c)

Using Newton’s first law, explain why the horizontal resultant force on the trolley is zero when it moves at constant velocity.

[1]
(d)

State the Newton’s third-law force pair corresponding to the frictional force exerted by the track on the trolley.

[2]

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