Question

A cart rolls down a ramp. The cart has a mass of 45 lb m . The cart starts at rest. The ramp is angled 0.17 radians downward from the horizontal. The cart travels a total distance of 97 meters along the ramp on the planet earth. What is the velocity of the cart at the bottom of the ramp? Assume friction is negligible

Answer 1

To solve this problem, we can use the principle of conservation of energy, which states that the initial potential energy of the system (the cart on the ramp) is converted to kinetic energy at the bottom of the ramp. We can express this as:

mgh = (1/2)mv^2

where m is the mass of the cart, g is the acceleration due to gravity, h is the height of the ramp, and v is the velocity of the cart at the bottom of the ramp.

First, we need to calculate the height of the ramp:

h = dsin(theta) = 97sin(0.17) = 16.16 meters

where d is the total distance traveled along the ramp and theta is the angle of the ramp.

Next, we can plug in the values and solve for v:

459.8116.16 = (1/2)45v^2

v^2 = (459.8116.16)/(1/2*45) = 748.91

v = sqrt(748.91) = 27.37 m/s

Therefore, the velocity of the cart at the bottom of the ramp is 27.37 m/s.