#giancoli solution

Hello! Here is another problem from the Giancoli book. Chapter 6, # 25.

A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160g by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, and (e) the final speed of the load assuming it started from rest.

Alright, here are the givens:

m = 285-kg

d = 22.0 m

a = 0.160g

initial velocity = 0

Here I have drawn a little diagram:

PART A - Finding the tension (FT) in the cable.

We have values for mass and acceleration. We can use the net force and what we know about the force of gravity on the load to find the tension in the cable.

F(net) = ma = FT - Fg

Here, we are subtracting the force of gravity because it is in the downward direction and in this case downward is negative.

We know that Fg = mg therefore,

ma = FT - mg

FT = ma + mg

FT = (285kg)(0.160g) + (285kg)g

FT = (285kg)(0.160)(9.8 m/s2) + (285kg)(9.8 m/s2)

FT = 3.24 x 103 N

Alright! We are done with Part A.

PART B - Finding the Net Work on the Load

We know the net force. We can use that and the distance that the load has traveled because of that force to find the net work. Here, the net force is parallel to the displacement, and we can use it to find the net work.

F(net)= ma = 446.88 N

W(net) = F(net)d

W(net) = (446.88 N)(22.0 m) = 9.83 x 103 J

Yay! We have found the answer to part B!

PART C - Finding the Work Done by the Cable on the Load

To find the work done by the cable only, we need to use the force that the cable applied on the load. This is the force of tension on the cable. We know that the force of tension on the cable is 3.24 x 103 N. We are going to use the value that wasn't rounded. If you did the calculation, you should have a number 3239.88 N.

WT = FTd = (3239.88 N)(22.0m) = 7.13 x 104 J

That was simple! Now, onto part D.

PART D - Finding the Work Done by Gravity on the Load

The work done by gravity on the load is simply the force of gravity on the load multiplied by the displacement. In this case, the work done is negative because the force is working against the movement of the load.

Wg = -Fgd = - mgd = - (285kg)(9.8m/s2)(22.0 m) = -6.14 x 104 J

PART E - Finding the Final Speed of the Load

We know that the initial velocity is zero.

We use the Work-Energy Principle

Wnet = KEf - KEi

Wnet = 0.5mvf2 - 0.5mvi2

Wnet = 0.5mvf2

We re-write the equation for v.

v = √(Wnet / 0.5 m) = √(9831.36 J / 142.5 kg) = 8.31 m/s

This problem is from the Giancoli Physics 6e text book. You can find it on Chapter 5. #33. I am aware that the answer is in the back of the book, but I feel it is important that others understand the process of getting to that answer. The book doesn't do this. Alrighty, here you go:

Two objects attract each other gravitationally with a force of 2.5 x 10-10 N when they are 0.25 m apart. Their total mass is 4.0 kg. Find their individual masses.

Alright, so here are our givens:

Fg = 2.5 x 10-10 N

d = 0.25 m

m1 + m2 = 4.0 kg

We can therefore say that:

m2 = 4.0kg - m1 and

m1 = 4.0kg - m2

We use the equation for Newton's Law of Universal Gravitation:

Fg = (G m1m2)/ d2

We solve the equation for m1m2

m1m2 = Fgd2 / G

After entering our values (remember to square the distance), we have the equation:

m1m2 = 0.23425kg

Since m1 = 4.0kg - m2

We can enter that equation into m1m2 = 0.23425kg

to get the equation:

m2(4.0kg - m2) = 0.23425kg

I will now stop using units (we know that the units are kg) because it makes it easier to see what I am doing.

Distribute the m2

4.0 m2 - m22 = 0.23425

We re-write as:

m22 - 4.0 m2 + 0.23425 = 0

We now have a quadratic equation that will give us two answers! These two answers will be the two masses.

Using the quadratic formula:

x = ( -b +/- √b2 - 4ac )/ 2a

m = (4.0 +/- √16 - 0.937) / 2

m = (4.0 +/- 3.8811) / 2

m = (4.0 + 3.8811) / 2     or       m = (4.0 - 3.8811) / 2

m = 3.94 kg   or m = 0.0595 kg

The final answers are: