# Force and Acceleration (2 Simple Prob.)

## Homework Statement

1. A sailboat with a mass of 2.0 x 10^3 kg experiences a tidal force of 3.0 x 10^3 N directed to the east and a wind force against its sails with a magnitude of 6.0 x 10^3 N directed toward the northwest (45 degrees N of W). What is the magnitude of the resultant acceleration of the boat?

A. 2.2 m/s^2
B. 2.1 m/s^2
C. 1.5 m/s^2
D. 4.4 m/s^2

2. An Olympic skier moving at 20.0 m/s down a 30.0 degree slope encounters a region of wet snow and slides 145 m before coming to a halt. What is the coefficient of friction between the skis and the snow?

A. 0.540
B. 0.740
C. 0.116
D. 0.470

1. a = F/m
2. ???

## The Attempt at a Solution

1. I drew a triangle with the 3000N and the 6000N, and found the other value by the Pythagorean Theorem to be 5196.15.
Then I found cos 45 = 0.707, and multiplied that by 5196.15 to get the force to be 3673.23.
I plugged that value in F=ma, and divided it by the mass of 2000 kg, and got 1.8.
What have I done wrong?
2. I've never encountered using distance with force. Should I find time, which would be 2900 seconds?

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

1. A sailboat with a mass of 2.0 x 10^3 kg experiences a tidal force of 3.0 x 10^3 N directed to the east and a wind force against its sails with a magnitude of 6.0 x 10^3 N directed toward the northwest (45 degrees N of W). What is the magnitude of the resultant acceleration of the boat?

A. 2.2 m/s^2
B. 2.1 m/s^2
C. 1.5 m/s^2
D. 4.4 m/s^2

2. An Olympic skier moving at 20.0 m/s down a 30.0 degree slope encounters a region of wet snow and slides 145 m before coming to a halt. What is the coefficient of friction between the skis and the snow?

A. 0.540
B. 0.740
C. 0.116
D. 0.470

1. a = F/m
2. ???

## The Attempt at a Solution

1. I drew a triangle with the 3000N and the 6000N, and found the other value by the Pythagorean Theorem to be 5196.15.
Then I found cos 45 = 0.707, and multiplied that by 5196.15 to get the force to be 3673.23.
I plugged that value in F=ma, and divided it by the mass of 2000 kg, and got 1.8.
What have I done wrong?
2. I've never encountered using distance with force. Should I find time, which would be 2900 seconds?
1. What do you mean by the 'other' value of 5196.15? 2. If you haven't learned work-energy methods yet, how about using the motion equation that relates velocity with distance and acceleration to solve for 'a' before then using Newton 2?

1. By the other value, I mean the third side of the triangle. Like 3000^2 + b^2 = 6000^2.
2. Which equation? final velocity = initial velocity + a(time) OR distance = initial velocity(t) + .5a(t)^2 ???
For the first equation, I got -0.0068. For the second equation, I plugged in the values and got -0.137 m/s^2 for the acceleration.
Even if I found the right acceleration, would what I do? I don't even have mass, and the mass formula involves acceleration.

PhanthomJay
Homework Helper
Gold Member
1. By the other value, I mean the third side of the triangle. Like 3000^2 + b^2 = 6000^2.
2. Which equation? final velocity = initial velocity + a(time) OR distance = initial velocity(t) + .5a(t)^2 ???
For the first equation, I got -0.0068. For the second equation, I plugged in the values and got -0.137 m/s^2 for the acceleration.
Even if I found the right acceleration, would what I do? I don't even have mass, and the mass formula involves acceleration.

In problem 1, you don't have a right triangle. You have 3000 to the right, and 6000 pointing upwards and left at 45 degrees. It is best to break each force into its x and y components, sum the x components together, sum the y components together, THEN use Pythagorus to solve the magnitude of the resultant. Draw a sketch and mind your plus and minus signs.

For Problem 2, you forgot v_f^2 =v_o^2 +2ad. solve for 'a'. Then draw the free body diagram of the skiier, identify the gravity and friction and other forces acting on him or her, and see what happens to 'm' in your equation .

1. THANKS! I got the right answer (hopefully) to be A, 2.2 m/s^2.

2. I solved that formula you gave me, and for acceleration I got a = 1.379 or 40/29.

If I do free body diagram, I only know theta of 30 degrees and gravity of 9.81 m/s^2. How will that help me?

PhanthomJay
Homework Helper
Gold Member
1. THANKS! I got the right answer (hopefully) to be A, 2.2 m/s^2.

2. I solved that formula you gave me, and for acceleration I got a = 1.379 or 40/29.

If I do free body diagram, I only know theta of 30 degrees and gravity of 9.81 m/s^2. How will that help me?
for number 1, please show your work, that answer is not correct. For problem 2, you've got to identify all the forces on the block....there's a bit of stuff involved here.. try a google search on 'blocks on inclined planes' for some help.

1. To find the third side of the triangle, I tried using the Law of Cosines. So angle A = 45 degrees, and the unknown side = a. b = 3000 and c = 6000.

a^2 = b^2 + c^2 - 2bc cos 45
a^2 = 3000^2 + 6000^2 - 2 (3000)(6000) cos 45
a^2 = 45000000 - 36000000 cos 45
a = 4420.877 N

So I plugged that into F=ma, so 4420.877 = 2.0 x 10^3 a, to get a = 2.2 m/s^2.

What have I done wrong?

2. Okay I will search it.

PhanthomJay
Homework Helper
Gold Member
1. To find the third side of the triangle, I tried using the Law of Cosines. So angle A = 45 degrees, and the unknown side = a. b = 3000 and c = 6000.

a^2 = b^2 + c^2 - 2bc cos 45
a^2 = 3000^2 + 6000^2 - 2 (3000)(6000) cos 45
a^2 = 45000000 - 36000000 cos 45
a = 4420.877 N

So I plugged that into F=ma, so 4420.877 = 2.0 x 10^3 a, to get a = 2.2 m/s^2.

What have I done wrong?

2. Okay I will search it.
The law of cosines is the court of last resort. I can't even remember the formula. Too easy too mess it up, as you have done. Didn't you like my hint to find components and use Mr. Pythagorus as your guide?

PhanthomJay