1) A block of mass M=120.0 kg is pushed along the floor with a force F=1400N whi

Question

1) A block of mass M=120.0 kg is pushed along the floor with a force F=1400N which makes an angle of 30degrees with the floor. The coefficient of fricton between the block and the floor is 0.3. Calculate the acceleration of the block.

2) an elecator and its passenger have a mass M=1400kg. the elevatoe is moving upward with an acceleration of 3m/s2. calculate the tension T in the cable as the elevator accelerates upward at 5m/s2.

4) an object starts with an initial velocity of 2.0 m/s and accelerates at 4.0 m/s2 for 6.0 seconds. it then decelerates at -2.0 m/s2 until its velocity is 4.0 m/s. the total distance covered is

Explanation / Answer

1) sum forces in the y

-1400*sin(30 degrees) + N - 120*9.81 = 0

N=1877.2

sum in the x

F cos 30 degrees - friction = ma

1400*cos(30 degrees) - 0.3*1877.2 = 120*a

a=5.4 m/s^2

2)

T - m g = ma

T = ma + mg = 1400*(9.81+3)= 17920 N

3)

sum forces

M1 g - friction = (m1 + m2) a

friction = u m2 g

75*9.81 - 0.3*140*9.81 = (75+140)*a

a= 1.51 m/s^2

4)

distance while acceleration
x = v0 t+ 1/2 a t^2 = 2*6 + 0.5*4*6^2= 84 m
v = v0 + a t = 2 + 4*6 = 26 m/s

then while slowing down
v^2 = v0^2 + 2 a x
4^2 = 26^2 -2*2*x
x=165

so total = 84 + 165 = 249 m

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