Question #6413

a box of mass 50kg is pulled on a incline 12m long and 2m high by the constant force of 100N from rest.it acquires velocity of 2m/s on reaching ,work done against frictional force is ???

Expert's answer

Let's consider all the forces acting on a box along an incline:

F1 = mg/sinα - the gravity force (here sinα = 2m/12m = 1/6);

F2 = 100N - pulling force;

F3 - unknown friction force.

Now let's write down the second Newton's law:

F1 + F2 - F3 = ma.

Let's guess what acceleration a is equal to.

S = V*t/2 ==> t = 2*S/V = 2*12/2 = 12 s (the moving time).

We know that

S = a*t²/2 ==> a = 2*S/t² = 2*12/12² = 2/12 = 1/6 m/s.

So,

F3 = F1 + F2 - ma = mg/sinα + 100 - ma = 50*9.8/6 + 100 - 50/6 = 520/3 = 173.(3) N.

F1 = mg/sinα - the gravity force (here sinα = 2m/12m = 1/6);

F2 = 100N - pulling force;

F3 - unknown friction force.

Now let's write down the second Newton's law:

F1 + F2 - F3 = ma.

Let's guess what acceleration a is equal to.

S = V*t/2 ==> t = 2*S/V = 2*12/2 = 12 s (the moving time).

We know that

S = a*t²/2 ==> a = 2*S/t² = 2*12/12² = 2/12 = 1/6 m/s.

So,

F3 = F1 + F2 - ma = mg/sinα + 100 - ma = 50*9.8/6 + 100 - 50/6 = 520/3 = 173.(3) N.

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