**PreQuiz 4.1 .:. Go Up**

**Dynamics**

**Show your work, and circle your answers and use sig figs
to receive full credit.**

1. A 362 gram object experiences an unbalanced force of 2.31 N. What is its acceleration? (6.38 m/s/s)

convert grams to kg:

F = ma

2.31 N = (0.362 kg)a

2. A 0.459 kg ball stops from +31 m/s in a distance of +0.17 m. What net force acted on the ball? What direction is the force? (-1297 = -1300 N (opposite original direction of ball))

Solve the suvat

s = 0.17 m

u = 31 m/s

v = 0 (stops)

a =

t =

Use v^{2}= u^{2}+ 2as:

0^{2}= (31 m/s)^{2}+2a(0.17 m)

a = -2826.47... m/s/s

F = ma = (0.459 kg)(-2826.47 m/s/s) = -1297.35 = -1300 N

Questions 1 and 2 are to see if you can solve a basic F = ma kind of problem. Problem 2 requires that you also solve a suvat problem to get the acceleration

3. Tycho exerts a force of 23 N sideways on a box and it moves at a constant speed. When he exerts 47 N of force, the box accelerates at 5.2 m/s/s in the direction he pushes. What is the box’s mass? (4.6 kg)

So if 23 N of force is what it takes to make it move at a constant rate, then there must be a friction force of 23 N opposing him. If the 47 N force is to the right then (+), and it is sliding to the right, then the 23 N friction force must be to the left, opposing him. So setting up Newton's second law gives:

<47 -23> = m(5.2)

m = 4.6153... = 4.6 kg

4. A 1217 kg elevator moving 5.2 m/s downward stops in 1.17 s. What must be the tension in the supporting cable as it is stopping? (17,347 = 17,000 N)

Solve the suvat

s =

u = -5.2 m/s

v = 0 (stops)

a = ?

t = 1.17 s

Use v = u + at: (Not in the data packet!!!!!!!)

0 = 5.2 + a(1.17)

a = +4.44444 m/s/s

The tension in the cable would not be the only force acting on the elevator - the force of gravity would also be acting downward with a force of F = ma, wt = mg, wt = (1217 kg)(9.81 N/kg) = 11938.77 N downwards (-). So setting up Newton's second law:

<T - 11938.77 N> = (1217 kg)(+4.444444 m/s/s)

T = 17347.7 N = 17,000 N with sig figs

Questions 3 and 4 are to see if you can solve a net force problem where there is more than one force acting on an object. Problem 4 is a vertical problem (weight is down = mg) with a suvat problem that you have to solve first, or last.

5. A 63.0 kg mass accelerates at 2.16 m/s/s down a 35.0^{o}
incline. What force must be acting along the plane for this to happen? What
direction is the force? (218 N up the plane)

This is like the net force problem above, except because the inclined plane is not vertical, only part of the weight (mg) acts down the plane. That part is mgSin(q) where q is the angle with the horizontal the plane makes. So we have a force of (63.0 kg)(9.81 N/kg)sin(35.0^{o}) = 354.49 N down (-) the plane. Now we are ready to set up Newton's second law using the convention that up the plane is positive, and down negative:

<F - 354.49 N> = (63.0 kg)(-2.16 m/s/s) - (The acceleration is negative because it isdownthe plane)

F = 218.4 N Up the plane (Since it is positive, it must be a forceupthe plane, had it been negative, it would have meant down the plane)

Question 4 is a problem on an inclined plane