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?
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.
How did you get -35 on #2?
35 N is opposing motion, so it provides a negative acceleration.
To answer these questions, we can use Newton's second law of motion and the equations of motion.
a) To find the force of friction, we first need to determine the net force acting on the toboggan. The net force is the vector sum of all forces acting on the toboggan. In this case, the only forces are the pull from the boys and the force of friction opposing the motion. The force of friction always acts in the direction opposite to the motion.
The force of friction can be found using the equation: Fnet = ma, where Fnet is the net force, m is the mass of the object, and a is the acceleration.
In this case, the mass of the toboggan is 30 kg, and the acceleration is given as 40 cm/s^2. However, we need to convert the acceleration to meters per second squared (m/s^2) to maintain consistent units. Since there are 100 centimeters in 1 meter, the acceleration is 0.4 m/s^2.
Fnet = ma
Fnet = (30 kg) * (0.4 m/s^2)
Fnet = 12 N
Now, we need to find the force of friction. Since the boys are pulling the toboggan with a force of 25 N each at an angle of 20 degrees, we can use trigonometry to find the horizontal component of their force.
The horizontal component is given by Fx = F * cos(theta), where F is the force and theta is the angle.
For each boy:
Fx = 25 N * cos(20 degrees)
Fx = 23.16 N
Since there are two boys pulling, the total horizontal force is:
Ftotal = 2 * Fx
Ftotal = 2 * 23.16 N
Ftotal = 46.32 N
Since the force of friction opposes the motion, it has the same magnitude as the net force but opposite direction:
Force of friction = - Fnet = - 12 N
Therefore, the force of friction acting on the toboggan is 12 N.
b) If the boys stop pulling after the toboggan reaches a speed of 3.0 m/s, we need to find the distance it will slide before coming to a stop. We can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s as the toboggan comes to a stop)
u = initial velocity (3.0 m/s)
s = distance traveled (to be found)
a = acceleration (-0.4 m/s^2 due to the force of friction opposing motion)
Rearranging the equation, we get:
0^2 = (3.0 m/s)^2 + 2(-0.4 m/s^2)s
0 = 9.0 m^2/s^2 - 0.8 m/s^2 * s
0.8 m/s^2 * s = 9.0 m^2/s^2
s = 9.0 m^2/s^2 / 0.8 m/s^2
s = 11.25 m
Therefore, the toboggan will slide a distance of 11.25 meters before coming to a stop.
c) To determine how much time it will take for the toboggan to slide this distance, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance traveled (11.25 m)
u = initial velocity (3.0 m/s)
a = acceleration (-0.4 m/s^2)
t = time (to be found)
Rearranging the equation, we get a quadratic equation:
0.4/2 * t^2 - 3.0t + 11.25 = 0
Solving this equation using the quadratic formula or factoring, we find that t ≈ 7.5 seconds.
Therefore, the toboggan will take approximately 7.5 seconds to slide a distance of 11.25 meters.