Before we go through the questions, we need to calculate and determine some values first.
r = 11.5 m
<span>m = 280 kg </span>
<span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N
</span>
1) What is the magnitude of the normal force on the care when it is at the bottom of the circle.
<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>
<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>
<span>Centripetal force = 7119.55 N </span>
<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>
<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>
<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>
√<span>(gr) </span>
√<span>(9.8 x 11.5) = 10.62 m/s</span>
Answer:
Explanation:
a)
Ff = μmgcosθ
Ff = 0.28(1600)(9.8)cos(-84)
Ff = 458.9217...
Ff = 460 N
b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.
W = Ffd
W = 458.9217(-49.4/sin(-84)
W = 22,795.6119...
W = 23 kJ
c) same assumptions as part b
The change in potential energy minus the work of friction will be kinetic energy.
KE = PE - W
½mv² = mgh - (μmgcosθ)d
v² = 2(gh - (μgcosθ)(h/sinθ))
v = √(2gh(1 - μcotθ))
v = √(2(9.8)(49.4)(1 - 0.28cot84))
v = 30.6552...
v = 31 m/s
Answer:
A. It does not exhibit projectile motion and follows a straight path down the ramp.
Its really hurts
Explanation:
Charge A and charge B are 2.2 m apart. Charge A is 1.0 C, and charge B is
2.0 C. Charge C, which is 2.0 C, is located between them and is in
electrostatic equilibrium. How far from charge A is charge C?
