A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?

Explanation:

kinetic energy was converted to potential energy in the spring.

the answer is in the above image

Explanation:

V=ds/dt

Where s =distance traveled

Given

V= 2ti - 3t²j

ds/dt=2ti - 3t²j

ds=(2ti - 3t²j)dt

Integrating

S= 2t²i/2 - 3t³j/3 + C

S=t²i - t³j + C

Since the object starts from Rest when t=0 and s(distance)=0

0=0²i - 0³j + C

C=0

Therefore

S=t²i - t³j

At t=2sec

S=(2)²i - (2)³j

S=4i - 8j

Magnitude of S(distance) = √4²+(-8)²

S= 4√5 meters

S=8.94meters.

2.mass =2kg

F(t)=5e^-0.1t

From Newton 2nd Law

F(t) = mdv/dt

5e^-0.1t= 2dv/dt

2dv = (5e^-0.1t)dt

Integrating

2V(t) = 5e^-0.1t/(-0.1) + C

2v(t) = - 50e^-0.1t + C

Since it started from rest at t=0. That is v(0)=0

2(0) = -50e^-0.1(0) + C

0 = -50 + C

C= 50

v(t) = -50e^-0.1t + 50

At t=10sec

v(10) = -50e^-0.1(10) + 50

V= -18.39 + 50

V= 31.61ms-¹.

While using your calc to evaluate -50e^-0.1(10)

Don't forget the -(minus) sign at the top of the exponential.

If you neglect it... You'll have a different answer.

Hope this helps.

Have a great day!!!