Find Coefficient of Friction for Toy Car Rolling Down Slope

In summary: Thank you so much for your help! In summary, the toy car of mass 80g rolls down a rough slope at an angle of 16 degrees to the horizontal, reaching a distance of 80cm before hitting a rubber barrier and bouncing back with its speed halved, reaching a height of 10cm. Using the equations of motion, we can solve for the coefficient of friction between the car and the slope to be 0.0956. This problem is from the CIE Mechanics 1 textbook by Cambridge.
  • #1
Shah 72
MHB
274
0
A toy car of mass 80g rolls from rest 80cm down a rough slope at an angle of 16 degree to the horizontal. When it hits a rubber barrier at the bottom of the slope it bounces back up the slope with its speed halved and reaches a height of 10 cm. Find the coefficient of friction between the car and the slope.
m= 0.08kg, u= 0m/s, s=0.8m, a>0
F=m×a
a1= 10 (sin 16- mu cos16)
V^2=u^2+2as
V^2=16(sin16-mu cos 16)
Coming up= a<0, s=0.1m
Speed is halved
So u^2=8(sin 16- mu cos 16)
a2=-10(sin 16+ mu cos 16)
V^2= u^2+2as
I got 0= 8( sin 16- mu cos 16)-2(sin16+mu cos 16)
I don't get the right ans given in the textbook which is coefficient of friction =0.0956
 
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  • #2
Shah 72 said:
A toy car of mass 80g rolls from rest 80cm down a rough slope at an angle of 16 degree to the horizontal. When it hits a rubber barrier at the bottom of the slope it bounces back up the slope with its speed halved and reaches a height of 10 cm. Find the coefficient of friction between the car and the slope.

10 cm is not the "height" ... it's the distance the toy travels back up the slope.
Is this exactly how the problem was stated?

downhill ...

$a_1 = g(\sin{\theta} - \mu \cos{\theta})$

$v_f^2 = 0^2 + 2a_1 \cdot (0.8)$

uphill ...

$a_2 = -g(\sin{\theta}+\mu \cos{\theta})$

$0^2 = \left(\dfrac{v_f}{2}\right)^2 + 2a_2 \cdot (0.1)$only unknowns are $v_f$ and $\mu$

solve the system
 
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  • #3
skeeter said:
10 cm is not the "height" ... it's the distance the toy travels back up the slope.
Is this exactly how the problem was stated?

downhill ...

$a_1 = g(\sin{\theta} - \mu \cos{\theta})$

$v_f^2 = 0^2 + 2a_1 \cdot (0.8)$

uphill ...

$a_2 = -g(\sin{\theta}+\mu \cos{\theta})$

$0^2 = \left(\dfrac{v_f}{2}\right)^2 + 2a_2 \cdot (0.1)$only unknowns are $v_f$ and $\mu$

solve the system
Yes that's exactly the way problem is stated.
 
  • #4
Shah 72 said:
Yes that's exactly the way problem is stated.
a1=10(sin16- mu cos 16)
V^2=0+2×10×0.8(sin 16- mu cos 16)
For going uphill
V=0m/s, u^2=4(sin 16- mu cos 16)
V^2= 4(sin 16- mu cos 16)- 2×10×0.1(sin16+ mu cos 16)
I still get the wrong ans for coefficient of friction
Textbook ans is 0.0956
 
Last edited:
  • #5
skeeter said:
downhill ...

$a_1 = g(\sin{\theta} - \mu \cos{\theta})$

$v_f^2 = 0^2 + 2a_1 \cdot (0.8)$
$v_f^2 = 16\sin{\theta} - 16\mu \cos{\theta}$

uphill ...

$a_2 = -g(\sin{\theta}+\mu \cos{\theta})$

$0^2 = \left(\dfrac{v_f}{2}\right)^2 + 2a_2 \cdot (0.1)$
$v_f^2 = 8\sin{\theta} + 8\mu\cos{\theta}$

$16\sin{\theta} - 16\mu \cos{\theta} = 8\sin{\theta} + 8\mu\cos{\theta}$

$2\sin{\theta} - 2\mu \cos{\theta} = \sin{\theta} + \mu\cos{\theta}$

$\sin{\theta} = 3\mu\cos{\theta}$

$\mu = \dfrac{\tan{\theta}}{3} = 0.0956$

please cite the source for these problems ...
 
  • #6
skeeter said:
$v_f^2 = 16\sin{\theta} - 16\mu \cos{\theta}$

$v_f^2 = 8\sin{\theta} + 8\mu\cos{\theta}$

$16\sin{\theta} - 16\mu \cos{\theta} = 8\sin{\theta} + 8\mu\cos{\theta}$

$2\sin{\theta} - 2\mu \cos{\theta} = \sin{\theta} + \mu\cos{\theta}$

$\sin{\theta} = 3\mu\cos{\theta}$

$\mu = \dfrac{\tan{\theta}}{3} = 0.0956$

please cite the source for these problems ...
Thank you so so so much!
 
  • #7
skeeter said:
$v_f^2 = 16\sin{\theta} - 16\mu \cos{\theta}$

$v_f^2 = 8\sin{\theta} + 8\mu\cos{\theta}$

$16\sin{\theta} - 16\mu \cos{\theta} = 8\sin{\theta} + 8\mu\cos{\theta}$

$2\sin{\theta} - 2\mu \cos{\theta} = \sin{\theta} + \mu\cos{\theta}$

$\sin{\theta} = 3\mu\cos{\theta}$

$\mu = \dfrac{\tan{\theta}}{3} = 0.0956$

please cite the source for these problems ...
It's from cie mechanics
 

Related to Find Coefficient of Friction for Toy Car Rolling Down Slope

1. What is the coefficient of friction?

The coefficient of friction is a measure of the amount of resistance between two surfaces when they come into contact with each other. It is a dimensionless number that ranges from 0 to 1, with 0 indicating no friction and 1 indicating maximum friction.

2. How is the coefficient of friction calculated?

The coefficient of friction is calculated by dividing the force of friction between two surfaces by the normal force pressing the surfaces together. This can be expressed in the equation: μ = F/N, where μ is the coefficient of friction, F is the force of friction, and N is the normal force.

3. Why is the coefficient of friction important for a toy car rolling down a slope?

The coefficient of friction is important for a toy car rolling down a slope because it determines the amount of friction between the wheels of the car and the surface of the slope. This friction affects the speed and movement of the car, and can also impact its ability to stay on the slope without slipping or sliding off.

4. How can the coefficient of friction be measured for a toy car rolling down a slope?

The coefficient of friction can be measured for a toy car rolling down a slope by conducting an experiment where the car is rolled down the slope at different angles and measuring the distance it travels. By using the equation μ = F/N and rearranging it to solve for μ, the coefficient of friction can be calculated using the measured values of F (force of friction) and N (normal force).

5. What factors can affect the coefficient of friction for a toy car rolling down a slope?

The coefficient of friction for a toy car rolling down a slope can be affected by factors such as the type of surface the car is rolling on, the weight and design of the car, the angle of the slope, and any external forces acting on the car (e.g. wind). Additionally, the presence of any lubricants or debris on the surface can also impact the coefficient of friction.

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