A box with an initial speed of 3 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 1/3 and an incline of ( pi )/3 . How far along the ramp will the box go?

2 Answers
Mar 17, 2018

Here,as the tendency of the block is to move upwards,hence the frictional force will act along with the component of its weight along the plane to decelerate its motion.

So,net force acting downwards along the plane is (mg sin ((pi)/3) + mu mg cos ((pi)/3))

So,net deceleration will be ((g sqrt(3))/2 + 1/3 g (1/2))=10.12 ms^-2

So,if it moves upward along the plane by xm then we can write,

0^2 =3^2 -2×10.12×x (using, v^2=u^2 -2as and after reaching maximum distance,velocity will become zero)

So, x=0.45m

Mar 17, 2018

The distance is =0.44m

Explanation:

Resolving in the direction up and parallel to the plane as positive ↗^+

The coefficient of kinetic friction is mu_k=F_r/N

Then the net force on the object is

F=-F_r-Wsintheta

=-F_r-mgsintheta

=-mu_kN-mgsintheta

=mmu_kgcostheta-mgsintheta

According to Newton's Second Law of Motion

F=m*a

Where a is the acceleration of the box

So

ma=-mu_kgcostheta-mgsintheta

a=-g(mu_kcostheta+sintheta)

The coefficient of kinetic friction is mu_k=1/3

The acceleration due to gravity is g=9.8ms^-2

The incline of the ramp is theta=1/3pi

The acceleration is a=-9.8*(1/3cos(1/3pi)+sin(1/3pi))

=-10.12ms^-2

The negative sign indicates a deceleration

Apply the equation of motion

v^2=u^2+2as

The initial velocity is u=3ms^-1

The final velocity is v=0

The acceleration is a=-10.12ms^-2

The distance is s=(v^2-u^2)/(2a)

=(0-9)/(-2*10.12)

=0.44m