In Figure P4.30, m1 = 10.0 kg and m2 = 4.5 kg. The coefficient of static friction between m1 and the horizontal surface is 0.60 while the coefficient of kinetic friction is 0.30.
Figure P4.30
(a) If the system is released from rest, what will its acceleration be?
1 m/s2
(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?
2 m/s2
To find the acceleration of the system in both scenarios, we need to consider the forces acting on each mass and apply Newton's second law of motion.
(a) If the system is released from rest, meaning both masses are initially stationary and there is no motion, the only force acting on m1 is the force of static friction exerted by the surface. The force of static friction can be calculated as the product of the coefficient of static friction (μs) and the normal force (which is equal to the weight of the object, mg).
The force of static friction on m1 is given by:
f_static = μs * N
where f_static is the force of static friction, μs is the coefficient of static friction, and N is the normal force.
Substituting the given values, we have:
f_static = 0.60 * (10.0 kg * 9.8 m/s²) = 58.8 N
Since the system is initially at rest, the force exerted by m2 must be equal to the force of static friction on m1 for the system to remain in equilibrium. Therefore, we can set up the following equation:
m2 * g = f_static
where m2 is the mass of object m2 and g is the acceleration due to gravity.
Substituting the given values, we have:
4.5 kg * 9.8 m/s² = 58.8 N
Now, we can calculate the net force acting on the system by subtracting the force of static friction from the force exerted by m2:
Net force = f_applied - f_static
Since there is no applied force in this scenario, the net force is equal to zero.
Using Newton's second law, F = ma, we can find the acceleration of the system:
Net force = m * a
0 = (10.0 kg + 4.5 kg) * a
0 = 14.5 kg * a
Therefore, the acceleration of the system in scenario (a) will be zero, or 0 m/s².
(b) If the system is set in motion with m2 moving downward, we need to consider both static and kinetic friction.
The force of static friction, as calculated earlier, is 58.8 N. Since the system is set in motion, m2 exerts a downward force greater than the force of static friction. This additional force is given by:
f_downward = m2 * g
Substituting the given values, we have:
f_downward = 4.5 kg * 9.8 m/s² = 44.1 N
The net force will then be the difference between the downward force and the force of static friction:
Net force = f_downward - f_static
Net force = 44.1 N - 58.8 N = -14.7 N
(Note that the negative sign indicates that the net force is in the opposite direction of motion).
Using the equation F = ma and substituting the net force, we can find the acceleration of the system:
-14.7 N = (10.0 kg + 4.5 kg) * a
-14.7 N = 14.5 kg * a
Solving the equation, we find:
a = -14.7 N / 14.5 kg = -1.01 m/s²
The acceleration of the system in scenario (b) will be -1.01 m/s². Since the negative sign indicates acceleration in the opposite direction of motion, we can take the absolute value to find the magnitude. Therefore, the acceleration of the system in scenario (b) will be approximately 1.01 m/s².