Question

A 3kg block slides along a floor with coefficient of kinetic friction k = 0.3, initially moving at 7.0m/s. It travels for 2.0 meters, then encounters a ramp sloped upward at 40o. The ramp also has a coefficient of kinetic friction k = 0.3.

a) How fast is the block moving when it reaches the bottom of the ramp?

b) How far up the ramp does the block slide, before momentarily coming to rest?

Answer 1

(a)

Given, Initial velocity of the block = 7 m/sec

Let's find the acceleration acting on the block. For that we need to find the net force acting on the block. The only force acting on the block is the frictional force, which is opposing the motion.

Frictional force on the horizontal surface = F_f = - mu_k * m * g

Negative because the force is acting against the direction of motion

Therefore, acceleration due to friction = F_f / m = -mu_k * g = - 0.3 * 9.81 = - 2.943 m/s²

Now, from v² - u² = 2 * a * s

v² - (7)² = 2 * -2.943 * 2

v² = 49 - 2 * 2.943 * 2

v² = 37.228

=> v = 6.1015 m/sec

The velocity of the block when it reaches the bottom of the ramp = 6.1015 m/sec

(b)

The above velocity will be the initial velocity after the block starts moving up with the ramp

Now, we have to find the net acceleration along the inclined surface.

Net, acceleration = F_net / m

F_net = Frictional force acting downwards along the incline + The Weight component along the incline acting downwards.

=> F_net = mu_k * mg * cos(theta) + mg * sing(theta)

Net acceleration along the inlined surface = F_net / m

=> a = mu_k * g * cos(theta) + g * sin(theta)

=> a = 0.3 * 9.81 * cos(40) + 9.81 * sin(40)

=> a = -8.56 m/s².

Negative because, the acceleration is opposing the motion.

Now, from v² - u² = 2 * a * s

v = 0 u = 6.1015, a = -8.56 m/s².

=> 0 - 6.1015² = 2 * -8.56 * s

=> s = 6.1015² / (2 * 8.56)

=> s = 2.1745 meters

Therefore, the block goes 2.1745 meters on the incline before momentarily coming to rest