Help with forces problem

  • #1
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Homework Statement


A 2.34-kg cart on a long, level, low-friction track is heading for a small electric fan at 0.24 m/s . The fan, which was initially off, is turned on. As the fan speeds up, the magnitude of the force it exerts on the cart is given by at^2, where a = 0.0200 N/s2.

A) What is the speed of the cart 3.5 s after the fan is turned on?
B) After how many seconds is the cart's velocity zero?

Homework Equations


F = ma
Vf = Vi + a*t

The Attempt at a Solution


So for part a I first found the force of the fan on the cart : (0.02)(3.5)^2 = 0.245 N. Then I found the force of the cart, but to do that I first found the acceleration: 0.24 = 0 + a(3.5) where "a" ended up being 0.06857 m/s^2. I plugged this into F = ma to find the force of the cart: F = (2.34)(0.06857) = 0.160457 N. I subtracted the force of the fan from the force of the cart: 0.160457 N - 0.245 N = - 0.0845 N. I then used this value to find the acceleration of the cart with the fan going against it: - 0.0845 N = (2.34 kg)a where "a" ended up being -0.036. I then plugged this back into the kinematic equation: Vf = (0.24 m/s) + (-0.036 m/s^2)(3.5) = 0.11 m/s . Not totally confident in what I did. Could you guide me as to where I went wrong? Thanks!
 

Answers and Replies

  • #2
billy_joule
Science Advisor
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The force decelerating the car isn't constant, or even linear so you'll need to use calculus to find the velocity of the car.
 
  • #3
TSny
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There is no "force of the cart" that acts on the cart. There is only one horizontal force acting on the cart which is the force that the fan exerts on the cart.

Note that the formula Vf = Vi + a*t is for an object moving with constant acceleration.
 
  • #4
38
3
There is no "force of the cart" that acts on the cart. There is only one horizontal force acting on the cart which is the force that the fan exerts on the cart.

Note that the formula Vf = Vi + a*t is for an object moving with constant acceleration.
So how do you take into account the force of the fan on the cart? Because the acceleration changes with time so you couldn't just use the acceleration of the fan minus the acceleration of the cart right?
 
  • #5
SammyS
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Homework Statement


A 2.34-kg cart on a long, level, low-friction track is heading for a small electric fan at 0.24 m/s . The fan, which was initially off, is turned on. As the fan speeds up, the magnitude of the force it exerts on the cart is given by at^2, where a = 0.0200 N/s2.

A) What is the speed of the cart 3.5 s after the fan is turned on?
B) After how many seconds is the cart's velocity zero?

Homework Equations


F = ma
Vf = Vi + a*t

The Attempt at a Solution


So for part a I first found the force of the fan on the cart : (0.02)(3.5)^2 = 0.245 N. Then I found the force of the cart, but to do that I first found the acceleration: 0.24 = 0 + a(3.5) where "a" ended up being 0.06857 m/s^2. I plugged this into F = ma to find the force of the cart: F = (2.34)(0.06857) = 0.160457 N. I subtracted the force of the fan from the force of the cart: 0.160457 N - 0.245 N = - 0.0845 N. I then used this value to find the acceleration of the cart with the fan going against it: - 0.0845 N = (2.34 kg)a where "a" ended up being -0.036. I then plugged this back into the kinematic equation: Vf = (0.24 m/s) + (-0.036 m/s^2)(3.5) = 0.11 m/s . Not totally confident in what I did. Could you guide me as to where I went wrong? Thanks!
The equation, Vf = Vi + a*t, is for constant acceleration. The acceleration in this problem is not constant.

It's unfortunate that the problem statement uses the variable a as the coefficient in the expression for force. I suggest that you use the time derivative of velocity, dv/dt, to denote acceleration. You will need to use calculus to solve this problem anyway.
 
  • #6
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So how do you take into account the force of the fan on the cart? Because the acceleration changes with time so you couldn't just use the acceleration of the fan minus the acceleration of the cart right?
Who says that the fan is accelerating? Have you tried drawing a free body diagram yet, or do you feel that you have advanced beyond the need to draw free body diagrams?
 
  • #7
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3
Who says that the fan is accelerating? Have you tried drawing a free body diagram yet, or do you feel that you have advanced beyond the need to draw free body diagrams?
I'm talking about the air coming from the fan. Which the problem describes as "As the fan speeds up, the magnitude of the force it exerts on the cart is given by at^2, where a = 0.0200 N/s2", which I assumed meant the air coming from the fan was accelerating with respect to time. I did indeed draw a free body diagram
 
  • #8
20,869
4,546
I'm talking about the air coming from the fan. Which the problem describes as "As the fan speeds up, the magnitude of the force it exerts on the cart is given by at^2, where a = 0.0200 N/s2", which I assumed meant the air coming from the fan was accelerating with respect to time. I did indeed draw a free body diagram
Please write out the force balance equation you derived from your free body diagram.
 
  • #9
SammyS
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I'm talking about the air coming from the fan. Which the problem describes as "As the fan speeds up, the magnitude of the force it exerts on the cart is given by at^2, where a = 0.0200 N/s2", which I assumed meant the air coming from the fan was accelerating with respect to time. I did indeed draw a free body diagram
The air from the fan is only accelerating in the following sense. As time passes, the speed of the air exiting the fan is increasing. However, that's not the same as any particular air mass accelerating. This is all irrelevant to solving the problem anyway.

Follow Chet's suggestions.

Again, as I suggested, express acceleration in Newton's 2nd Law, as dv/dt .
 

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