Question

A block of mass $2.0kg$ at rest on an inclined plane of inclination ${37}^{\circ }$ is pulled up the plane by applying a constant force of $20N$ parallel to the incline. The force acts for one seconds. (a) Show that the work done by the applied force does not exceed $40J$. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is $40J$. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take $g=10m/s^2$.

Answer 1

Force F = 20N .

Mass m = 2.0 kg

Initial velocity u = 0

acceleration, a = 10m/s${}^2$

t = 1 sec.

Now Refer to the attachment, See the free body diagram of the block.

Force works on the block:-

Weight, W = mg

W = 2 $\times$ 10

W = 20N (which is Downward)

Normal force N = mg cos37

N = 20 $\times$ 0.80

N = 16 N. (perpendicular & upward to the plane )

Here Applied Force, P = 20N (which is down along the plane)

Now For Final Speed, We know the formula:-

v = u + at

v = 0 + 10 $\times$ 1

v = 10 m/s

the Distance travelled s = ut + 0.5 at$\times$t

s = 0 + 0.5 $\times 10\times 1\times 1$

s = 5 m.

Now,

(a) So work done by the force of gravity in 1 sec. = F $\times$ d

= 20 N $\times$ 5m

= 100 J.

(b) Here the weight act as downward, so distance travelled in downward.

= 5 $\times$ sin37

= 5 $\times$ 0.6

= 3 m.

so work done by gravity,

= 20 N $\times$ 3 m

= 60 J.

(c) Now, work done by all the forces

= change in Kinetic energy

= $\frac{1}{2}m(v^2-u^2)$

= $0.5\times 2.0\times ({10}^2-0^2)$

= 100 J.

W.D by frictional force

= work was done by all forces -( work was done by Normal force + work done by applied force + work done by gravity )

= 100 J - (100 + 60 +0 )

= 100 - 160

= -60 J