Answer to Question #193526 in Mechanics | Relativity for Gil Dror

Question

The diagram depicts a trajectory consisting of two segments: The first segment, AB, is a quarter of a particle circle with a radius of R = 45. Point B is on the circular trajectory. A second section, BD, is a non-smooth sloping slope whose angle is a = 30 degrees. Release from rest from point A a 2kg weight box stopping on the inclined plane at point D, which is 50m away from point B.

1. Calculate the speed of the box at point B. Show in the drawing the direction of the speed at point. necrosis. 

2. Given that beta = 60 degrees. Draw a diagram depicting the forces acting on the box at point K and calculate the normal force acting on the body at this point.

3. Calculate the frictional force acting on the box during its movement.

 4. Calculate the kinetic coefficient of friction between the slope and the box. 

5. Where during the movement of the box will the magnitude of the normal force acting on it be maximum? Necrosis.

6. What will be the work of the normal force on the box during its movement from the point you found in section 5 to its stopping? Necrosis.

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Answer 1

Solution.

$R=45m;$

$m=2kg;$

$\alpha=30^o;$

$\beta=60^o;$

$l=50m;$

$1. mgR=\dfrac{m{v_b}^2}{2}\implies v_b^2=2gR; v_b=\sqrt{2gR};$

$v_b=\sqrt{2\sdot10ms^{-2}\sdot45m}=30m/s;$

$2. N=\sqrt{m^2g^2cos^2\beta+m^2g^2sin^2\beta}=mg;$

$N=2kg\sdot 10m/s^2=20N;$

$3. A_f=\dfrac{mv_b^2}{2}-mglsin\alpha;$

$A_f=\dfrac{2kg\sdot(30m/s)^2}{2}-2kg\sdot10m/s^2\sdot50m\sdot 0.5=400J;$

$A_f=F_fl\implies F_f=\dfrac{A_f}{l};$

$F_f=\dfrac{400J}{50m}=8N;$

$4. F_f=\mu N=\mu mgcos\alpha\implies \mu=\dfrac{F_f}{mgcos\alpha};$

$\mu=\dfrac{8N}{2kg\sdot10m/s^2\sdot 0.87}=0.46;$

$5.$ The magnitude of the normal force will be maximum at point B.

$6.$  The operation of normal force while moving from point B to the stop will be zero.

Answer: $1. v_b=30m/s;$

$2. N=20N;$

$3. F_f=8N;$

$4. \mu=0.46;$

$5.$ The magnitude of the normal force will be maximum at point B;

$6.$ The operation of normal force while moving from point B to the stop will be zero.