Question

A cyclist comes to a skidding stop at 50 m. During this process, the force on the cycle due

to the road is 1000 N and is directed opposite to the motion. How much work does the

road do on the cycle?​

Answer 1

Explanation:

work done=force *distance

given that

d=50,f=1000N

so : wd=f*d

wd=50*1000

wd=50000, or , 5*10^4Joules.

Answer 2

Given:

• Force on cycle by road = 1000 N
• Displacement = 50 metres

To find:

• Net work done by road ?

Calculation:

First of all, the question clearly mention that due to opposing force of the road on the cycle, the cyclist comes to a stop.

This means that the force vector and displacement vector are opposite to one another. So, the angle between them is 180°.

Now, work will be :

$W = f \times d \times \cos( \theta)$

$\implies \: W = 1000 \times 50 \times \cos( {180}^{ \circ} )$

$\implies \: W = 1000 \times 50 \times ( - 1)$

$\implies \: W = - 50000 \: joule$

$\implies \: W = - 50 \: kilojoule$

So, the work done by the road on cycle is -50 kJ.