# Physics 1303 ```Physics 1309
Name:
Quiz 3
1. Strings A and B are tied to a vertical string that
supports a weight. The tension in string A is 100 N.
Calculate the tension in string B.
30&ordm;
A
60&ordm;
B
B cos(60) – A cos(30) = 0
0.500 B – 0.866 A = 0
B = 0.8660/0.500 A = 1.73 A = 173 N
2. A chemistry student is driving a car with mass equal to 1000 kg at 23.0 m/s on an icy
road. She suddenly slams on her brakes and begins to skid. Assuming the road is
perfectly level and the coefficient of sliding friction between her tires and the ice equals
0.135, how far does the car slide before coming to a complete stop?
a = - 0.135 * 9.8 m/s2 = - 1.323 m/s2
0 = v2 + 2a x  x = v2 / -2a = (23.0 m/s)2 / 2*1.323 m/s2 = 200 m
3. A driver is driving a 1000 kg car over a hill. The radius of curvature of the hill equals
80 m and the speed of the car equals 20 m/s.
a. Make a FBD for the car when
it is at the top of the hill. Show
all the forces acting on the car.
N
mg
b. Calculate the centripetal acceleration of the car when it is at the top of the hill.
a = v2 / R = (20 m/s)2 / 80 m = 5.0 m/s2
c. Calculate the normal force exerted on the car by the road when the car is at the top of
the hill.
mg – N = m a  9800 N – N = 5000 N  –N = 9800 N –5000 N = 4800 N
4. A 200 kg box is at rest on a rough horizontal surface. The coefficient of kinetic friction
between the box and surface equals 0.25. A biology student, applying a 590 N horizontal
force to the box, moves the box 10 m.
a. Calculate the work done by the student.
WS = 590 N * 10 m = 5900 J
b. Calculate the work done by friction.
f = - 0.25 mg = - 0.25 * 200 kg * 9.8 m/s2 = - 490 N
Wf = - 490 N * 10 m = -4900 J
c. Calculate the change in the kinetic energy of the box.
Wnet = WS + Wf = 5900 J – 4900 J = 1000 J
5. An athlete throws 0.145 kg baseball so that at its highest point, it is 10 meters above
the ground and its speed equals 20 m/s.
a. Calculate the work done by gravity as the ball moves from its highest point to the point
where it hits the ground.
W = mgy = 0.145 kg * 9.8 m/s2 * 10 m = 14.2 J
b. Calculate the kinetic energy of the ball just before it hits the ground.
KE = +14.2 J
KEinitial = &frac12; * 0.145* (20 m/s)2 = 29.0 J
KEfinal = 29.0 J + 14.2 J = 43.2 J
```