Solve questions A, B, and C B and C are in the bottom Class Management | Help HW

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Solve questions A, B, and C B and C are in the bottom Class Management | Help HW07 Begin Date: 10/12/2015 1200:00 AM-Du th mass ms e Date: 10/16/2015 11:5900 PM End Date: 12/22/2015 11:59 00 PM m,- 43 kg is standing at the top of a ramp which is h-4.5 m above the ground (10%) Problem S: A skateboarder with The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vy- 6.8 m/s Randomized Variables ms= 43 kg = 4.5 m v,= 6.8 m/s 33% Part (a) write an expression for the work. done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. Grade Summary Deductions 0% Potential 100% cos(0) sin(6) Submissions Attenpts remaining 0% per attempt) 112 detailed view ms Submit Hint I give up Hints: deduction per hint. Hints remaining Feedback: St deduction per feedback 33% Part (b) The ramp makes an angle with the ground where 30 degrees Write an expression for the magnitude of the fiction force between the ramp and the skateboarder 33% Part (c) when the skateboarder reaches the bottom of the ramp, he continues moving with the speed vronto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force. F grass in newtons, between the skateboarder and the grass

Explanation / Answer

here,

mass of the skateboarder , ms = 43 kg

hy = 4.5 m

final velocity , vf = 6.8 m/s

(a)

the work done by the friction force , Wf = ms*g*hy - ( 0.5 * ms * vf^2)

Wf = 43 * 9.8 * 4.5 - 0.5 * 43 * 6.8^2

Wf = 20.98 J

the work done by the friction force is 20.98 J

the expression of the work done by the friction force is (ms*g*hy - ( 0.5 * ms * vf^2))

(b)

theta = 30 degree

work done by the friction force = friction force * hy/sin(30)

ms*g*hy - ( 0.5 * ms * vf^2) = friction force * 2*hy

friction force = (ms*g*hy - ( 0.5 * ms * vf^2))/(2*hy)

the expression of the friction force is ((ms*g*hy - ( 0.5 * ms * vf^2))/(2*hy))

(c)

initial speed , v0 = 6.8 m/s

distance , d = 5 m

let the accelration be a

using third equation of motion

v^2 - v0^2 = 2 * a*d

0 - 6.8^2 = 2 * a * 5

a = 4.62 m/s^2

frictional force , ff = m*a

ff = 198.66 N

the friction force between grass and the skateboard is 198.66 N

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