Really simple problem - can't get right answer

  • Thread starter justagirl
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In summary, the conversation is about a 50kg block sliding on a horizontal surface with a coefficient of kinetic friction of 0.60. A force of 400 N is applied at an angle of 20° above the horizontal. The question asks for the magnitude of the block's acceleration, with answer choices of 0.54 m/s2, 2.31 m/s2, 3.27 m/s2, 6.78 m/s2, and 8.11 m/s2. After calculations, it is determined that the acceleration in the x-direction is 1.67 m/s2, but the y-component of the force (400*sin(20)) also needs to be taken into account.
  • #1
justagirl
29
0
Suppose a 50kg block slides along a
horizontal surface where the coefficient of kinetic friction
between the block and the surface is uk = 0.60. A force F =
400 N is now applied, where the
angle of the force above horizontal is 20°.

What is the magnitude of the acceleration of the block?

0.54 m/s2
2.31 m/s2
3.27 m/s2
6.78 m/s2
8.11 m/s2

Here's what i did:

F(x-net) = F(applied) - F(friction) = ma
a = F(applied) - F(friction) / m
= (400)(cos 20) - (50 * 9.8 * 0.6) / 50
= 375 N - 294 N / 50 kg
= 1.67 m/s2

I checked my numbers quite a few times and keep getting
1.67, which is not one of the answer choices. The only thing
I saw was that 3.72 m/s2 is exactly double what I got, but I
can't find an explanation for that.

Thanks!
 
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  • #2
The answer is 3.27.
You gave the right anwser for the acceleration in the x-direction. Now when you look vertically in the y direction you still have the y-component of the 400 N-force that you forgot (gravity and normal force annihilate each other)

This y-component = 400*sin(20)

Then calculate a_y and you already have a_x =1.67 which is correct.

The magnitude is sqrt((a_x)² + (a_y)²)

marlon
 
  • #3


It seems like you have made a small calculation error in your solution. When finding the net force in the x-direction, you should use the sine function instead of the cosine function since the applied force is at an angle of 20° above the horizontal. So, the correct calculation would be:

F(x-net) = F(applied) - F(friction) = ma
a = F(applied) - F(friction) / m
= (400)(sin 20) - (50 * 9.8 * 0.6) / 50
= 136 N - 294 N / 50 kg
= -3.16 m/s2

Since the direction of the acceleration is opposite to the direction of the applied force, the magnitude of the acceleration would be 3.16 m/s2. This value is closer to the answer choice of 3.27 m/s2, so it is likely the correct answer.

It's always a good idea to double check your calculations and make sure you are using the correct trigonometric function for the given angle. I hope this helps you find the correct answer!
 

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