a wooden block with a mass of 2.0kg starts from at the top of an inclined plane and ends with a force of 25N at the bottom what was the rate of acceleration of the block

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≅