Question

A 4.75 kg block is sent up a ramp inclined at an angle ????=31.5° from the horizontal. It is given an initial velocity ????0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is ????k=0.368 and the coefficient of static friction is ????s=0.663. How far up the ramp in the direction along the ramp does the block go before it comes to a stop?

Answer 1

Answer:

d = 13.7 m

Explanation:

When block is moving upwards along the inclined plane

then the block is decelerated due to gravity as well as due to friction and speed of the block by which it is projected upwards is given

$v_i = 15 m/s$

deceleration caused to the block due to net force opposite to the motion is given as

$F = - mg sin\theta - \mu mgcos\theta$

$a = \frac{F}{m}$

$a = -g(sin\theta + \mu cos\theta)$

since block is sliding on the inclined plane

so here we can say that the coefficient of the friction must be kinetic friction here

$a = -9.81(sin31.5 + 0.368 cos31.5)$

$a = - 8.2 m/s^2$

now for finding the distance upto which it will stop is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 15^2 = 2(-8.2) d$

$d = 13.7 m$