Question

Q4. Consider the skier on a slope shown in the figure below. Her mass including equipment is 55.0 kg.

Answer 1

Answer:

Part a)

When there is no friction then acceleration is

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then acceleration is

$a = 3.33 m/s^2$

Explanation:

Part a)

As we know that the skier is on inclined plane

So here if there is no friction then net force along the inclined plane is given as

$F = mg sin\theta$

now acceleration of the skier is given as

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

$a = g sin\theta$

$a = 9.81(sin25)$

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then net force along the inclined plane is given as

$F = mg sin\theta - F_f$

now acceleration of the skier is given as

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

$a = g sin\theta - \frac{F_f}{m}$

$a = 9.81(sin25) - \frac{45}{55}$

$a = 3.33 m/s^2$