F of magnituda 50.0 N acts on it. The coefficient of kinetic friction between bl

Question

F of magnituda 50.0 N acts on it. The coefficient of kinetic friction between block and plane is 0.300. what are the (a) magritude and (b) direction (up or down the plane) In the figure, a 5.00 kg black is sent sliding up a plana inclined at 37.0e whille a horizontal force of the block's acceleration? The block's initial speed is 4.20 m/s. (c) How far up the plane does the block go? (d) When it reaches its highest point, does it remain at rest or slide back down the plane? (a) N (b) Tdown (c) Number-3.11 dremain at rest

Explanation / Answer

here,

mass , m = 5 kg

theta = 37 degree

F = 50 N

uk = 0.3

a)

the accelration , a = net force /effective mass

a = ( F * cos(theta) - m * g * sin(theta) - uk * ( m * g * cos(theta) + F * sin(theta))) /m

a = ( 50 * cos(37) - 5 * 9.81 * sin(37) - 0.3 * ( 5 * 9.81 * cos(37) + 50 * sin(37))) /5 m/s^2

a = - 2.1 m/s^2

the magnitude of accelration is 2.1 m/s^2

b)

acting downwards

c)

initial speed , u = 4.2 m/s

the distance travelled , s = u^2 /(2 a)

s = 4.2^2 /(2 * 2.1) = 4.2 m

it will go 4.2 m

d)

at maximum height , it remains at rest

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