Answer:
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 × 10
W = 20N (which is Downward)
Normal force N = mg cos37
N = 20 × 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 × 1
v = 10 m/s
the Distance travelled s = ut + 0.5 at×t
s = 0 + 0.5 ×10×1×1
s = 5 m.
Now,
(a) So work done by the force of gravity in 1 sec. = F × d
⇒20 N × 5m
⇒100 J.
(b) Here the weight act as downward, so distance travelled in downward.
⇒5 × sin37
⇒5 × 0.6
⇒3 m.
so work done by gravity,
⇒20 N × 3 m
⇒60 J.
(c) Now, work done by all the forces
⇒change in Kinetic energy
⇒
2
1 m(v2
−u 2
)
⇒0.5×2.0×(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 - 1
Hope this helps you≅
