Answer:
d = 69 .57 meter
Explanation:
First case
Speed of car ( v ) = 20.5 mi/h = 9.164 M/S
distance ( d ) = 11.6 meter ( m = mass of the car )
Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J
Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N
Second case
v = 50.2 mi/h = 22.44135 m/s
d = ?
Work done = 0.5 * 22.44² * m J = 251.7768 * m J
Since the braking force remains the same .
3.619 m = ( 251.7768 m / d )
d = 69 .57 meter
For the part a) we need only the momentum of the box and we have the data to find it.
Momentum is given by,

where clearly, p is the momentum, m the mass of the box and v is the velocity.
Substituting,

For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,

<em>It is the same that part a)</em>
Freezing.................
Answer:
14 N
Explanation:
The tension in the second string is puling both the masses of 20 kg and 8 kg with acceleration of 0.5 m s⁻²
So tension in the second string = total mass x acceleration
= 28 x .5 = 14 N . Ans..