Question #5360

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

Expert's answer

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.

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.

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