Question

A box of mass m = 10 kg is projected up an inclined plane from its foot with
a speed of 20 m/s as shown in the figure. The coefficient of friction u between
the box and the plane is 0.5. Find the distance travelled
by the box on the plane before it stops for the first time.​

Answer 1

Answer:

The distance traveled  by the box on the plane before it stops for the first time is d=20.38m

Explanation:

From the Newton's second law we get F=ma

In the diagram the term N=Normal Reaction, f= friction, W=weight=mg

Friction Coefficient=μ

The second figure is plotted on x-y plane. So the weight on x plane is Wx=mg.sinθ

Wy=mg.cosθ

So along x axis -Wx-f=ma .......(1) and along y axis we get,

N-Wy=0

=> N=mg.cosθ

In the equation no 1,

=> -mg.sinθ-μmg.cosθ=ma

=> a=-g(sinθ+μmg)= -9.81 m/s²

Let assume that the velocity is V and the distance is d

Now,

Using Kinematics,

Vi=20m/s

Vf=0

So, Vf²=Vi²+2ad

=> 20²-(2x9.81xd)= 0

=> d=20²/2x9.81 =20.38 m

Answer 2

Answer:

20 m

Explanation: