Question

A 28-kg rock approaches the foot of a hill with a speed of 15 m/s. This hill slopes upward at a constant angle of 40.0∘
above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20, respectively.
(a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock.
(b) Will the rock remain at rest at its highest point, or will it slide back down the hill?
(c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

Answer 1

Answer:

Part a)

$H = 9.27 m$

Part B)

$\mu_s = 0.84$

since the required static coefficient is more then the given value so it will not remain at rest at highest position.

Part c)

$v_f = 11.76 m/s$

Explanation:

Frictional force on the rock while it is moving upwards along the plane is given as

$F_k = \mu_k mg cos\theta$

now we have

$F_k = 0.20 (28 \times 9.8)cos40$

$F_k = 42 N$

now by energy conservation we can say

Work done by friction = loss in mechanical energy

$F_k s = (\frac{1}{2}mv^2) - (mgH)$

here we know that

$H = s sin\theta$

$F_k \times (\frac{H}{sin\theta}) = \frac{1}{2}(28)(15^2) - (28 \times 9.8\times H)$

$42 \times \frac{H}{sin40} = 3150 - 274.4H$

$65.3H = 3150 - 274.4 H$

$H = 9.27 m$

Part B)

Now at the maximum height position we can say that force due to static friction must be balanced by its weight along the plane

so we will have

$mg sin\theta = \mu_s mg cos\theta$

$tan\theta = \mu_s$

$tan40 = \mu_s$

$\mu_s = 0.84$

since the required static coefficient is more then the given value so it will not remain at rest at highest position.

Part c)

Again by mechanical energy conservation law

Work done by friction = loss in mechanical energy

$F_k (2s) = (\frac{1}{2}mv_i^2) - (\frac{1}{2}mv_f^2)$

here we know that

$H = s sin\theta$

$F_k \times (2\frac{H}{sin\theta}) = \frac{1}{2}(28)(15^2) - \frac{1}{2}(28)(v_f^2)$

$42 \times 2\frac{9.27}{sin40} = 3150 - 14v_f^2$

$1211.4 = 3150 - 14v_f^2$

$v_f = 11.76 m/s$