Question

Gayle runs at a speed of 4.15 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 15.0 m?

Answer 1

Answer:

$v=7.87m/s$

Explanation:

To complete the question so: Gayle's mass is  49.0  k g , the sled has a mass of  5.15  k g  and her brother has a mass of  30.0 k g .

$K_1+E_1+E_2=K_{f1}+K_{f2}+K_{f3}$

$\frac{1}{2}*m_1*v_1+m_2*g*h_1+m_3*g*h_3=\frac{1}{2}*m_1*v_{f}^2+\frac{1}{2}*m_2*v_{f}^2+\frac{1}{2}*m_3*v_{f}^2$

$\frac{1}{2}*m_1*v_1+m_2*g*h_1+m_3*g*h_3=\frac{1}{2}*(m_1+m_2+m_3)*v_f^2$

Solve to find their speed at the bottom if the hill so:

$\frac{1}{2}*49kg*(4.15m/s)^2+5.15kg*9.8m/s^2*15.0+30kg*9.8m/s^2*5.0m=\frac{1}{2}*(49kg+5.15kg+30kg)v^2$

$2649.001=42.075kg*v^2$

$v^2=61.964m^2/s^2$

$v=7.87m/s$