Question

# A gardener pushes a lawn roller through a distance $20m$. If he applies a force of $20kg-w$t in a direction inclined at ${60}^{o}$ to the ground, the work done by him is

A

1960 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

196 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

1.96 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

196 KJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 1960 JStep1: Given dataA gardener pushes a lawn roller through a distance $S=20m$. If he applies a force of $20kg-w$t in a direction inclined at $\theta ={60}^{o}$ to the ground.Step2: Formula used$W=FS\mathrm{cos}\theta \left[whereW=workdone,F=cons\mathrm{tan}tforceapplied,S=dispalcement,\theta =anglebetweenFandS\right]$Step3: Calculating the work doneLet us write down the given data, force applied by the gardener to push the lawn roller, $F=20kg-wt\phantom{\rule{0ex}{0ex}}⇒1kg-wt=9.8N$Therefore, $F=20kg-wt\phantom{\rule{0ex}{0ex}}F=20×9.8N\phantom{\rule{0ex}{0ex}}F=196N$Displacement of the lawn roller, $S=20m$The angle between the direction of force applied and direction of displacement is, $\theta ={60}^{o}$As the force applied by the gardener is constant, we can use the formula, $W=FS\mathrm{cos}\theta$ to find out the work done by the gardener.Substituting $F=196N,S=20m,and\theta ={60}^{o}$ in the formula we get,$W=196×20×\mathrm{cos}{60}^{\circ }\phantom{\rule{0ex}{0ex}}W=196×20×\frac{1}{2}\left[As\mathrm{cos}{60}^{\circ }=\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}\therefore W=1960J$The work done by the gardener in pushing the lawn roller is $1960J$.Hence, option A is the correct answer.

Suggest Corrections
4
Explore more