1. A 31kg child slides down a playground slide at a constant speed . The slide h
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Question
1. A 31kg child slides down a playground slide at aconstant speed. The slide has a height of 3.6mand is 7.4m long. Using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child.
2. A 1500kg car is approaching the hill shown in (Figure 1)at 12m/s when it suddenly runs out of gas. Neglect any friction.What is the car's speed after coasting down the other side?
3. An ice cube of mass 50.0g can slide without friction up and down a 25.0degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 0.100m . The spring constant is 25.0N/m . When the ice cube is released, how far will it travel up the slope before reversing direction?
A) Identify the initial and final gravitational potential energies.
B) Identify the initial and final elastic potential energies.
(Enter your answers, separated by a comma, in terms of some or all of the variables m, k, x, d, ?, and the acceleration due to gravity, g.)
1. A 31kg child slides down a playground slide at aconstant speed. The slide has a height of 3.6mand is 7.4m long. Using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child. 2. A 1500kg car is approaching the hill shown in (Figure 1)at 12m/s when it suddenly runs out of gas. Neglect any friction.What is the car's speed after coasting down the other side? 3. An ice cube of mass 50.0g can slide without friction up and down a 25.0degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 0.100m . The spring constant is 25.0N/m . When the ice cube is released, how far will it travel up the slope before reversing direction? A) Identify the initial and final gravitational potential energies. B) Identify the initial and final elastic potential energies. (Enter your answers, separated by a comma, in terms of some or all of the variables m, k, x, d, ?, and the acceleration due to gravity, g.)Explanation / Answer
1)
m*g*sintheta*L - f*L = 0
f = m*g*sintheta = 31*9.8*h/L = (31*9.8*3.6)/7.4 = 147.79 N
2) o.5*M*Vi^2 = -M*g*h + 0.5*M*Vf^2
Vf = sqrt(Vi^2 + 2*g*h) = sqrt((12*12)+(2*9.8*5)) = 15.56 m/s
3) M*g*sin25*L = 0.5*K*x^2
L = 0.603 m
B) Ui = 0.5*K*x^2
Uf = 0
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