Final answer:
The normal force on the shoebox is 14.7 N, the frictional force is 2.205 N, the acceleration of the shoebox is 0.86333 m/s², the speed of the shoebox after 2.0 seconds is 1.7267 m/s, and the distance traveled by the shoebox after 3.0 seconds is 3.878 m.
Step-by-step explanation:
(a) What is the Normal Force?
To find the normal force, we first need to understand that the normal force is the force exerted by a surface perpendicular to the object. In this case, the normal force is equal to the weight of the shoebox, which can be calculated using the equation F_n = m*g, where m is the mass of the shoebox (1.5 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, the normal force is 1.5 kg * 9.8 m/s^2 = 14.7 N.
(b) What is the Magnitude of the Frictional Force?
The magnitude of the frictional force can be calculated using the equation F_f = u * F_n, where u is the coefficient of friction (0.15) and F_n is the normal force. Substituting in the known values, we get F_f = 0.15 * 14.7 N = 2.205 N.
(c) What is the Acceleration of the shoebox?
To find the acceleration of the shoebox, we can use the equation F_net = m*a, where F_net is the net force, m is the mass of the shoebox, and a is the acceleration. In this case, the net force is equal to the force from the rocket minus the force of friction. So, F_net = 3.5 N - 2.205 N = 1.295 N. Substituting in the known values, we get 1.295 N = 1.5 kg * a. solving for a, a = 1.295 N / 1.5 kg = 0.86333 m/s^2.
(d) How fast is the shoebox traveling after 2.0 seconds?
To find the final velocity of the shoebox after 2.0 seconds, we can use the equation v_f = v_i + a*t, where v_i is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is assumed to be 0 m/s (since the shoebox starts from rest), the acceleration is 0.86333 m/s^2, and the time is 2.0 seconds. Substituting in the known values, we get v_f = 0 m/s + 0.86333 m/s^2 * 2.0 s = 1.7267 m/s.
(e) How far has the shoebox traveled after 3.0 seconds?
To find the distance traveled by the shoebox after 3.0 seconds, we can use the equation d = v_i*t + 0.5*a*t^2, where d is the distance, v_i is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s, the acceleration is 0.86333 m/s^2, and the time is 3.0 seconds. Substituting in the known values, we get d = 0 + 0.5*0.86333 m/s^2 * (3.0 s)^2 = 3.878 m.