A 450 g block is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.670, and the coefficient of kinetic friction is 0.380. A force of magnitude P pushes the block forward and downward as shown in the figure below. Assume the force is applied at an angle of 39.0° below the horizontal.
(a) Assuming P is large enough to make the block move, find the acceleration of the block as a function of P. (Use P as necessary and round numbers to the third decimal place.)
a(P) = m/s2
(b) If P = 6.81 N, find the acceleration and the friction force exerted on the block.
a =
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2
Ffriction = N
(c) If P = 13.62 N, find the acceleration and the friction force exerted on the block.
a = m/s2
Friction = N
(d) Describe in words how the acceleration depends on P.
force P is adding to the normal force. Figure that first.
normalforce=mg+Psin39
friction force= mu(mg+Psin39)
Pcos39-mu(mg+Psin39)=ma solve for a
To solve this problem, we can begin by considering the forces acting on the block. These forces include the force of gravity, the normal force exerted by the surface, and the frictional force.
(a) The first step is to calculate the gravitational force acting on the block. The formula for gravitational force is given by F_gravity = m * g, where "m" is the mass of the block and "g" is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, the mass of the block is 450 g, which can be converted to kilograms by dividing by 1000. Thus, the mass becomes 0.45 kg, and the gravitational force is F_gravity = 0.45 kg * 9.8 m/s^2.
Next, we need to determine the maximum frictional force that can be exerted on the block before it starts to move. The formula for the maximum static friction force is given by F_static_max = μ_static * N, where μ_static is the coefficient of static friction and N is the normal force. The normal force can be calculated as N = m * g, which in this case is 0.45 kg * 9.8 m/s^2.
Since the block is initially at rest, the applied force P must overcome the maximum static friction force in order to set the block in motion. Thus, we have the inequality P > F_static_max. Substituting the expressions for F_static_max and N, we have P > μ_static * m * g.
Once the block starts to move, the frictional force changes from static to kinetic friction. The formula for kinetic friction is given by F_kinetic = μ_kinetic * N, where μ_kinetic is the coefficient of kinetic friction.
To find the acceleration of the block, we need to use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the difference between the applied force and the frictional force, so we have P - F_kinetic = m * a.
Now, let's solve for the acceleration of the block as a function of P:
P - μ_kinetic * m * g = m * a
(a) a(P) = (P - μ_kinetic * m * g) / m
(b) To find the acceleration and the friction force when P = 6.81 N, we can substitute this value into the equation obtained in part (a) and solve for a. Additionally, we can substitute the value of a obtained into the equation F_kinetic = μ_kinetic * N to calculate the friction force.
(c) Similarly, when P = 13.62 N, we can substitute this value into the equation obtained in part (a) and solve for a. Then, substitute the value of a obtained into the equation for the friction force.
(d) The acceleration is directly proportional to the applied force P. As the force increases, the acceleration of the block also increases.