Mechanics ...Get the box

Classical Mechanics Level 1

A box weighing 15 N slides down from the top of an inclined plane making an angle of 30 degree with the horizontal.Calculate the work done by the force of gravity in moving the box 5 m along the plane.Take g=10 m/s^2 & assume the force of friction negligible?

DETAILS- Give answer in joules like -if answer is 28.7 joules ,write 28.7

The answer is 37.5.

3 solutions

Krishna Karthik
Nov 19, 2018

Simple problem.

Acceleration= $gsin(30º)=5m/s^2$

$F_{incline}$ = $mgsin(30º)$

$F_{incline}=5m$

$F_{weight}=10m=15N$

$m=1.5kg$

$F_{incline}=5m=7.5$

Work= $Fd$ (no need of integration, constant force)

Work= $7.5*5=37.5J$

Mihir Chakravarti
Dec 13, 2014

The force along incline will be $mgsin30$ and the distance is $5 m$ along the inclined plane so work done by gravity is $7.5×5$ which is 37.5

Kartik Sharma
May 22, 2014

First, consider that the box is at the top of the plane. Then,

height(h1) would be = (hypo)(sin30)

let hypotenuse be x

h1 = xsin30 = x/2

In the same way when it has covered 5 m

h2 = sin30(x-5) = x-5/2

now,

By W-E theorem, Work = KE2 - KE1 (here it would be U2 or U1 as in this case PE = KE)

= mgh2 - mgh1

= mg(h2 - h1) = 15(x-5/2 - x/2)

= 15(-5/2) = -75/2 = -37.5

we get a -ve work as we are reducing the height of the object.

Mechanics ...Get the box

The answer is 37.5.

3 solutions

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