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Physics Mechanics- friction

Shah 72

Member
Apr 14, 2021
193
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

Well-known member
MHB Math Helper
Mar 1, 2012
935
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
 
Last edited:

Shah 72

Member
Apr 14, 2021
193
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

Member
Apr 14, 2021
193
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:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
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

Member
Apr 14, 2021
193
$v_f^2 = 16\sin{\theta} - 16\mu \cos{\theta}$




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

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

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

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

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

please cite the source for these problems ...
Thank you so so so much!!!
 

Shah 72

Member
Apr 14, 2021
193
$v_f^2 = 16\sin{\theta} - 16\mu \cos{\theta}$




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

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

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

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

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

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