4. A UPS shipping center worker accidentally releases a 200 kg crate that was be

Question

4. A UPS shipping center worker accidentally releases a 200 kg crate that was being held at rest at the top of a ramp that is 3.5 m long and inclined at 40 degrees to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is 0.25. a. How fast is the crate moving as it reaches the bottom of the ramp? b. How far will it subsequently slide across the floor? c. Do the answers to (a) and (b) increase, decrease, or remain the same if we halve the mass of the crate?

Explanation / Answer

4)
a) let a is the acceleration of the on th inlined surface

Fnet = m*g*sin(40) - mue_k*N

m*a = m*g*sin(40) - mue_k*m*g*cos(40)

a = g*sin(40) - mue_k*g*cos(40)

= 9.8*sin(40) - 0.25*9.8*cos(40)

= 4.42 m/s^2

now Apply, v^2 - u^2 = 2*a*d

v^2 - 0^2 = 2*4.42*3.5

v = sqrt(2*4.42*3.5)

= 5.56 m/s <<<<<<------------Answer

b) accelration on horozonatl plane

a' = -g*mue_k

= -9.8*0.25

= -2.45 m/s^2

now Apply, (v'^2 - v^2) = 2*a'*d'

==> d' = (v'^2 - v^2)/(2*a)

= (0^2 - 5.56^2)/(2*(-2.45))

= 6.3 m <<<<<<------------Answer

c) ramain the same.<<<<<<------------Answer

beacuse these values do not depend on mass.

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