Question

1. A woman has a mass of 63.0kg .

A. What is her weight on earth?

B. What is her mass on the moon, where g=1.62m/s2?

C. What is her weight on the moon, where g=1.62m/s2?

2.

A. How much force does an 72kg astronaut exert on his chair while sitting at rest on the launch pad?

B. How much force does the astronaut exert on his chair while accelerating straight up at 14m/s2 ?

3.

A. A 4000 kg truck is parked on a 15? slope. How big is the friction force on the truck?

B. A 1700kg car traveling at a speed of 36m/s skids to a halt on wet concrete where ?k = 0.20. How long are the skid marks?

4.

A. What is the drag force on a 1.6 m wide, 1.4 m high car traveling at 10 m/s (? 22 mph)?

B. What is the drag force on a 1.6 m wide, 1.4 m high car traveling at 30 m/s (? 65 mph)?

5. A 1000 kg car pushes a 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N.

A. What is the magnitude of the force of the car on the truck?

B. What is the magnitude of the force of the truck on the car?

Answer 1

1) weight = mass * gravitation(g)

So A) weight = 63 * 9.8 = 617N

B) mass is same every where. Weight that changes according to the gravitatinal acceleration.

so her mass on the moon is 63Kg.

C) Weight On moon = 63 * 1.62 = 102.06N

2)

A) Launch pad is at rest. So force astronaut exerts on the chair is his weight, that is equal to 72 * 9.8 = 705.6N

b) Launch pad moving up at acceleration of 14 m/s2. So the astronaut exerts a force of 72 * (9.8 + 14) = 1713.6N.

3) A). As the truck is in equlibrium(not moving on the slope), the frictional force is balanced by the force of the truck along the slope.

frictional force = $mg *sin\Theta$ = 4000 * 9.8 * sin 15 = 4000 * 9.8 * 0.2588 = 10145N.

B) kinetic frictional energy = kinetic energy of the car

$\mu$_k*m*g *d = (1/2) * m*v²

d = v² / (2*$\mu$_k*g)

d = 1296 / 3.92

d = 330.6 m

4) A) Drag force= (1/4)(density of air)(Area)(v^2)

D= (1/4)(1.2kg/m^3)(1.4mx1.6m)(10 m/s)^2
D= 67.2N

B. Drag force= (1/4)(density of air)(Area)(v^2)

D= (1/4)(1.2kg/m^3)(1.4mx1.6m)(30 m/s)^2
D= 604.8

5) Assuming that there is movement, and thus acceleration, leads to

F = ma

In this case m = m car + m truck; a is assumed constant and not need for the answer.

F = F accelerate the car + F accelerate the truck.
a is the same for both, so the F accelerating the car is