A 10 kg cart coasts up and over a frictionless path of radius 2.08 m. as it pass

Question

A 10 kg cart coasts up and over a frictionless path of radius 2.08 m. as it passes the top of its speed is 3.2 m/s.

a) What is the magnitude and direction of centripetal force on the cart at this point?

b) What is the magnitude and direction of the gravitational force on the cart at this point?

c) What two forces add together to produce the centripetal force on the cart?

d) What is the normal force (magnitude and direction) on the cart from the surface of the hill at the top of the hill?

e) Will the cart stay on the surface of the hill or become airborne?

Explanation / Answer

a) F=mv^2/r=10kg*3.2m/s^2/2.08m so F=49 N b)Fg=m*g=10kg*9.8m/s^2=98N c)Normal force and gravity d)Fc=mg+Fn=49N=98N+Fn Fn=-49N (depending on axis choice) e) The cart will be airborne

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