An empty sled of mass 25 kg slides down a muddy hill with a constant speed of 2.4 m/s. The slope of the hill is inclined at an angle of 15 degrees with the horizontal.
a. Calculate the frictional force on the sled as it slides down the slope.
b. Calculate the coefficient of friction between the sled and the muddy surface of the slope.
c. The sled reaches the bottom of the slope and continues on horizontal ground. Assume the same coefficient of friction. In terms of velocity and acceleration, describe the motion of the sled as it travels on horizontal ground.
The correct answer and explanation is :
a. Frictional Force on the Sled
To calculate the frictional force on the sled, we need to recognize that the sled is moving with a constant velocity, meaning the net force acting on the sled along the slope is zero. This implies that the force due to gravity (parallel to the slope) is exactly balanced by the frictional force.
- Calculate the component of gravitational force parallel to the slope: The gravitational force acting on the sled is given by ( F_g = mg ), where ( m ) is the mass of the sled and ( g ) is the acceleration due to gravity (( 9.8 \, \text{m/s}^2 )). The component of gravitational force parallel to the slope is: [
F_{\parallel} = mg \sin(\theta)
] Where:
- ( m = 25 \, \text{kg} )
- ( g = 9.8 \, \text{m/s}^2 )
- ( \theta = 15^\circ ) Now, calculating ( F_{\parallel} ): [
F_{\parallel} = (25)(9.8) \sin(15^\circ)
]
[
F_{\parallel} = 245 \times 0.2588 = 63.44 \, \text{N}
]
- Frictional force: Since the sled moves at constant speed, the frictional force ( F_{\text{friction}} ) must balance the parallel component of gravity: [
F_{\text{friction}} = F_{\parallel} = 63.44 \, \text{N}
]
Thus, the frictional force on the sled is 63.44 N.
b. Coefficient of Friction
The frictional force is related to the normal force by the equation:
[
F_{\text{friction}} = \mu F_{\text{normal}}
]
Where:
- ( \mu ) is the coefficient of friction
- ( F_{\text{normal}} ) is the normal force (the force perpendicular to the surface)
- Calculate the normal force: The normal force ( F_{\text{normal}} ) is the component of the gravitational force perpendicular to the slope, which is: [
F_{\text{normal}} = mg \cos(\theta)
] Substituting the known values: [
F_{\text{normal}} = (25)(9.8) \cos(15^\circ)
]
[
F_{\text{normal}} = 245 \times 0.9659 = 236.74 \, \text{N}
] - Solve for the coefficient of friction ( \mu ): From the equation ( F_{\text{friction}} = \mu F_{\text{normal}} ), we can solve for ( \mu ): [
\mu = \frac{F_{\text{friction}}}{F_{\text{normal}}}
]
[
\mu = \frac{63.44}{236.74} = 0.268
]
Thus, the coefficient of friction between the sled and the surface is 0.268.
c. Motion on Horizontal Ground
When the sled reaches the bottom of the slope and continues on horizontal ground, the only horizontal force acting on it is the frictional force. This frictional force will cause the sled to decelerate until it comes to rest.
- Velocity and Acceleration on Horizontal Ground: On horizontal ground, the sled’s velocity will begin to decrease due to the opposing frictional force. The acceleration (deceleration) is given by Newton’s second law: [
F_{\text{friction}} = ma
] Where ( a ) is the acceleration (negative because it’s decelerating). From the earlier calculation, the frictional force is 63.44 N. The mass of the sled is 25 kg, so: [
63.44 = 25a
] Solving for ( a ): [
a = \frac{63.44}{25} = 2.54 \, \text{m/s}^2
] Therefore, the sled will decelerate at a rate of ( 2.54 \, \text{m/s}^2 ). - Behavior of the Sled:
- Initially, the sled is moving with a velocity of 2.4 m/s (the velocity at the bottom of the slope).
- The sled will continue to slow down due to the frictional force until it eventually comes to rest.
- The velocity will decrease over time, and the sled will come to a stop after a certain distance, which can be calculated using the kinematic equation: [
v^2 = u^2 + 2ad
] Where: - ( v ) is the final velocity (0 m/s when the sled stops),
- ( u ) is the initial velocity (2.4 m/s),
- ( a ) is the acceleration (-2.54 m/s², negative because it’s decelerating),
- ( d ) is the distance traveled before coming to rest. Solving for ( d ): [
0 = (2.4)^2 + 2(-2.54)d
]
[
0 = 5.76 – 5.08d
]
[
d = \frac{5.76}{5.08} \approx 1.13 \, \text{m}
]
Thus, the sled will decelerate with an acceleration of ( 2.54 \, \text{m/s}^2 ) and come to rest after traveling about 1.13 meters on horizontal ground.