Find Coefficient of Friction for Toy Car Rolling Down Slope

[Shah 72]

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

 

[skeeter]

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

 

[Shah 72]

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.

 

[Shah 72]

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

 

[skeeter]

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 ...

 

[Shah 72]

$v_f^2 = 16\sin{\theta} - 16\mu \cos{\theta}$

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

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

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

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

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

please cite the source for these problems ...

Thank you so so so much!

 

[Shah 72]

$v_f^2 = 16\sin{\theta} - 16\mu \cos{\theta}$

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

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

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

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

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

please cite the source for these problems ...

It's from cie mechanics