.012 J is going to be your answer
The answer is 0.000824653J
You need to use the formula Mass * Velocity^2 over 2
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity
d= 250s
t= 17s
a= d/t
=

= 4.7
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration.
The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N.
a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r.
The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec.
So a_N = 114 m/sec^2.
g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
Answer:
955.5N
Explanation:
The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

mass m = 65kg
radius of the loop r = 4m
velocity v = ?
g = 9.8 m/s²
To find the centripetal force, you need to find the velocity of the car at the top of the loop.
Use energy conservation:

At the top of the hill:

At the top of the loop:

Setting both energies equal and canceling the mass m gives:

Solving for v:

Using v in the first equation:
