1 ) A 2.50 kg box is moving to the right with speed 9.00 m/s on a horizontal, fr
ID: 1652100 • Letter: 1
Question
1 ) A 2.50 kg box is moving to the right with speed 9.00 m/s on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t)=( 6.00 N/s2 )t2
a) What distance does the box move from its position at t=0 before its speed is reduced to zero? Express your answer with the appropriate units.
b) If the force continues to be applied, what is the velocity of the box at 3.50 s ? Express your answer with the appropriate units.
2) A box with mass 14.0 kg moves on a ramp that is inclined at an angle of 55.0 above the horizontal. The coefficient of kinetic friction between the box and the ramp surface isk = 0.300.
a) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to the ramp surface and directed down the ramp, moving the box down the ramp
b) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0
that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
Explanation / Answer
1) Given,
m = 2.5 kg ; v0 = 9 m/s at t = 0 ;
F(t) = -6 N/s^2 t^2 (negative sinve its directed to left)
a)F = ma => a = F/m
a(t) = -(6/2.5) t^2 = -2.4 t^2
a = dv/dt => dv = a dt
dv = (-2.4 t^2) dt
on integrating we get
v = 2.4/3 t^3 + C = -0.8t^3 + C
at t = 0, v = 9 m/s
C = 9 m/s
v(t) = 9 - 0.8 t^3
v = dx/dt => dx = v dt
dx = (9 - 0.8t^3)dt
on integrating we get
x = 9t - 0.2 t^4
a)for t = 0 till it stops, v = 0
v = 9 - 0.8 t^3 = 0 => t = 2.24 s
x = 9 x 2.24 - 0.2 x 2.24^4 = 15.125 m
Hence, x = 15.125 m
b)v(3.5) = 9 - 0.8 x 3.5^3 = -25.3 m/s
Hence, v = -25.3 m/s
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