Question

the length of a smooth inclined plane of inclination 30degree is 5m the work done in moving a 10kg mass from the bottom of the inclined plane to top is

Answer 1

Answer:

Explanation:

Work = mglsin@

= 10×9.8×5×sin30°

= 245J

Answer 2

The work done in moving a 10kg mass from the bottom of the inclined plane to top is -250 J

Given : The length of the smooth inclined plane is 5 m

           Inclination of  the inclined plane is 30 degree

        Mass of the object is 10 kg

To Find : The work done in moving a 10 kg mass from the bottom of the inclined plane to top

Solution :  The work done in moving a 10kg mass from the bottom of the inclined plane to top is -250 J

It is given that the the length of a smooth inclined plane of inclination 30 degree is 5m

We have to find the work done in moving a 10kg mass from the bottom of the inclined plane to top

Now Work done is equal to

W = F × d Cos(theta)

Here f is is force

d is displacement

and theta is the angle between force and displacement

Now since the box is at the bottom of the inclined plane so the horizontal component of the force is mgSin(theta)

As shown in the figure  below angle between force and displacement tis 180 degree

So work done is

m×g×Sin(theta)×d×Cos(theta)

= 10×10×$\frac{1}{2}$×5×(-1)

= -250 J

So the work done in moving a 10kg mass from the bottom of the inclined plane to top is -250 J