How Does Ramp Angle and Friction Affect Ice Block Speed?

[physicsgirlie26]

A 8.20-kg block of ice, released from rest at the top of a 1.43-m-long frictionless ramp, slides downhill, reaching a speed of 2.11 m/s at the bottom.

What is the angle between the ramp and the horizontal?

What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 11.0 N parallel to the surface of the ramp?


I think I use the equations of ax=gsin(theta) and Fn=mgcos(theta)? I'm not sure what to do. Can someone please help me?

Thank you!

 

[Astronuc]

With the acceleration component down the ramp, then find the distance traveled to achieve a velocity of 2.11 m/s. It starts at rest.

In the second part, the friction works against the weight component acting down the ramp, so the acceleration is less. Find that acceleration and solve as one does for the first part of the problem.

 

[m2003]

To find the angle, we can use the equation for the acceleration of an object sliding down a ramp: a = gsin(theta). Rearranging this equation, we get theta = sin^-1(a/g), where a is the acceleration (in this case, 2.11 m/s^2) and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get theta = 12.7 degrees.

To find the speed of the ice at the bottom with the friction force, we can use the equation for the work done by friction: W = mgh - Ff*d, where W is the work done, m is the mass of the ice (8.20 kg), g is the acceleration due to gravity, h is the height of the ramp (1.43 m), Ff is the friction force (11.0 N), and d is the distance traveled (1.43 m). Rearranging this equation, we get v = sqrt((2mgh - 2Ff*d)/m), where v is the speed of the ice at the bottom. Plugging in the values, we get v = 1.95 m/s.