A 1927 kg car has a speed of 10.3 m/s when it hits a tree. The tree doesn’t move and the car comes to rest. Find the change in kinetic energy of the car.
Answer in units of J.
Find the amount of work done by the car as its front is pushed in.
Answer in units of J.
Find the magnitude of the force that pushed the front of the car in by 48 cm .
Answer in units of N
1
Expert's answer
2011-12-01T08:47:52-0500
Let's make following denotations: M = 1927 kg; V0 = 10.3 m/s; L = 0.48 m.
The energy of a car was E0 = (M*V0²)/2 = (1927*10.3²)/2 = 102217.715 J. and became E1 = 0. So, the change in kinetic energy is E0-E1 = 102217.715 - 0 = 102217.715 J.
Find the amount of work done by the car as its front is pushed in. Answer in units of J. The amount of work is equal to the kinetic energy change: A = E0-E1 = 102217.715 J. & Find the magnitude of the force that pushed the front of the car in by 48 cm. Answer in units of N
Let's find the acceleration of the car while it stopped: a = (V0 - V1)/T. Here T = 2*L/V0 is the time car to stop after hit and V1 = 0. So, a = (V0 - V1)*V0/(2*L) = 10.3²/(0.48*2) =& 110.51 m/s². Now let's use the second Newton's law: F = M*a = 1927*110.51 = 212953.57 N.
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
Leave a comment