Question

A 64.0-kg skier starts from rest at the top of a ski slope of height 62.0 m.

Answer 1

Answer:

Part a)

$v = 29.5 m/s$

Part b)

$v_f = 16.5 m/s$

Part c)

$F = 2915 N$

Explanation:

Part a)

As per work energy theorem we can say that work done by all forces on the ski must be equal to change in kinetic energy

so we will have

$W_g + W_{friction} = \frac{1}{2}m(v_f^2 - v_i^2)$

now we will have

$mgH - 1.10 \times 10^4 = \frac{1}{2}(64)(v^2 - 0)$

$64(9.80)(62) - 1.10 \times 10^4 = 32 v^2$

$v = 29.5 m/s$

Part b)

Now when the ski mover through a rough path then again we can use work energy theorem to find the final speed after crossing the patch

$W_f + W_{air} = \frac{1}{2}m(v_f^2 - v_i^2)$

$-F_{air} d - \mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)$

$-(160)(70) - 0.18(64\times 9.8)(70) = \frac{1}{2}(64)(v_f^2 - 29.5^2)$

$-19102.72 = 32(v_f^2 - 29.5^2)$

$v_f = 16.5 m/s$

Part c)

As the ski strike with snow drift then after moving a distance of d = 3.0 m the ski will stop

so here we can say that work done while it penetrate is equal to change in kinetic energy

$-F. d = \frac{1}{2}m(v_f^2 - v_i^2)$

$- F. 3 = \frac{1}{2}(64)(0 - 16.5^2)$

$F = 2915 N$