Consider a block on a table. This block is pushed by a spring attached to the wa

Question

Consider a block on a table. This block is pushed by a spring attached to the wall, slides across the table, and falls to the ground. The block has a mass m = 1.35kg. The spring constant is k = 560N/m, and the spring has been compressed by 0.11 m. The block slides a distance d = 0.65m across the table of height h = 0.75m. Assume the table is frictionless. What speed will the block have when it hits the floor? Assume the coefficient of kinetic friction between the block and table is mu_k = 0.16. What speed will the block have when it hits the floor? A 12.0&g; crate sits on the floor. A man wants to load it onto a truck by sliding it up a ramp 2.5m long, inclined at 30^degree. A worker giving no thought to friction, calculates that he can get the crate up the ramp by giving it an initial speed of 5.0m/s at the bottom and letting it go. But the friction is not negligible; the crate slides 1.6m up the ramp, stops and slides back down. Assuming that the friction force on the crate is constant, find its magnitude. How fast is the crate moving when it reaches the bottom of the ramp?

Explanation / Answer

here,

mass , m = 12 kg

length of ramp , l = 2.5 m

theta = 30 degree

initial speed , u = 5 m/s

a)

let the magnitude of frictional force be ff

s = 1.6 m

v^2 - u^2 = - 2 * s * ( g * sin(theta) + ff/m )

0 - 5^2 = - 2 * 1.6 * ( 9.8 * sin(30) + ff/12)

solving for ff

ff = 34.95 N

the magnitude of frictional force is 34.95 N

b)

let the speed when it reaches the bottom be v'

v^2 - u^2 = 2 * s * ( g * sin(theta) - ff/m )

v'^2 = 2 * 1.6 * ( 9.8 * sin(30) - 34.95/12)

solving for v'

v' = 2.52 m/s

the speed at the bottom of track is 2.52 m/s

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