Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:

Explanation:
Set up force equation
∑F=ma
∑F=W+FB
The minus sign for downward direction
I would say its a positive cgarge
Answer:
43.58 m
Explanation:
If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees
Using trigonometry ratio
Sin 5 = opposite/hypothenus
Where the hypothenus = 500m
Opposite = height h
Sin 5 = h/500
Cross multiply
500 × sin 5 = h
h = 500 × 0.08715
h = 43.58m
Therefore, the height above the starting point is equal to 43.58m
Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

hence, B_o = 1.013μT
The solution to your problem is as follows:
2.2Kg*9.8m/s = 21.56N
<span>
21.56N*1.25m = 26.95J </span>
<span>We're only concerned with the work done against gravity, lifting the books to 1.25 meters. the distance walked has no effect on the problem, unless you take into account the wind resistance and the force needed to overcome it. Also, lowering the books onto the shelf doesnt count, because gravity does the work on the books.</span>