A block of wood is projected up an inclined plane, starting at t = 0 at the bott

Question

A block of wood is projected up an inclined plane, starting at t = 0 at the bottom of the plane, with an initial speed of 6 m/s. The plane is inclined at 20 degree above the horizontal. Let the coefficient of kinetic friction be mu = 0.2 between block and plane. a) Apply Newton's second law to find the time taken for the block to travel 2 in tip along the plane, and find the velocity of the block at this point. b) Find the time and position (i.e. distance from bottom of plane) at which the velocity of the block has decreased to one-quarter its initial value.

Explanation / Answer

here,

initial speed , u = 6 m/s

theta = 20 degree

let the accelration of the block be a

using Newton's seccond law

m*a = (m * g * sin(theta) + u *m* g * cos(theta))

a = ( 9.81 * sin(20) + 0.2 * 9.81 * cos(20)) m/s^2

a = 5.2 m/s^2 downwards

a)

for s = 2 m

s = u * t + 0.5 * a * t^2

2= 6 * t - 0.5 * 5.2 * t^2

t = 0.4 s

v = u + a * t

v = 6 - 5.2 * 0.4

v = 3.92 m/s

b)

when v = u/4 = 1.5 m/s

let the time taken be t

v = u + a * t

1.5 = 6 - 5.2 * t

t = 0.87 s

the distance travelled , s = u * t + 0.5 * g * t^2

s = 6 * 0.87 - 0.5* 5.2 * 0.87^2

s = 3.25 m

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