Question

25 kg crate starting from rest at the top slides down a plane that makes an angle of 30 degree with the horizontal when it reaches the bottom of the 10 metre long slide its velocity is
8 metre per second the work done by the force of friction is closest to value of

Answer 1

Given:

25 kg crate starting from rest at the top slides down a plane that makes an angle of 30 degree with the horizontal when it reaches the bottom of the 10 metre long slide its velocity is 8 metre per second.

To find:

Work done by friction.

Calculation:

We shall apply WORK-ENERGY THEOREM:

• Work-Energy theorem states that the total work done by all the forces will be equal to the change in kinetic energy.

$\therefore \: W_{gravity} + W_{friction} = \Delta KE$

$= > \: mgh + W_{friction} = \dfrac{1}{2} m {v}^{2} - 0$

$= > \: mgh + W_{friction} = \dfrac{1}{2} m {v}^{2}$

$= > \: \{25 \times 10 \times 10 \sin( {30}^{ \circ} ) \} + W_{friction} = (\dfrac{1}{2} \times 25 \times {8}^{2} )$

$= > \: 1250 + W_{friction} = 800$

$= > \: W_{friction} = 800 - 1250$

$= > \: W_{friction} = - 450 \: joule$

So, work done by friction is -450 J.

HOPE IT HELPS.