Question

Another toboggan with a mass of 30kg is being accelerated at 40 cm/s^2 by two boys pulling ropes at an angle of 20degrees with the direction of travel. They are each pulling with a force of 25N.

a) what is the force of friction?

b) if, after reaching a speed of 3.0m/s, the boys stop pulling, how far will the toboggan slide?

c) how much will this slide take?

Answer 1

1.
F=ma
m=30kg
a=0.4 m/s²
F=30*0.4
= 12 N
Net force on the tobaggan in the direction of motion
= 2*25N*cos(20°)
= 47 N
Friction force = 47 N - 12N = 35 N

2. Use the equations of kinematics:
Equations of Kinematics
F = force of friction = 35 N
m = mass = 30 kg
a = acceleration
u = initial velocity = 3 m/s
v = final velocity
S = distance travelled
t = time
g = acceleration due to gravity

F = ma (frictional force)
-35N = 30kg * a
a = -35/30 = -1.17 m/s² (deceleration)
S = (v²-u²)/2a
= (0 - 3²)/(-2*1.17)
= 3.9 m

3. time?
v=u+at
0 = 3+(-1.17)t
t = 2.6 s.

Note: all numerical values are approximate. Please recheck values.

Answer 2

How did you get -35 on #2?

Answer 3

35 N is opposing motion, so it provides a negative acceleration.