Two horizontal forces, P and Q, are acting on a block that is placed on a table. We know that P = (1.10î + 4.25k)N but the magnitude and direction of Q are unknown. The block moves along the table. Assume there is no friction between the object and the table. Here the mass of the block is 4.50 kg.

(a) Determine the force Q when the block moves with constant velocity. Express your answer in vector form.
Q = N

(b) Determine the force Q when the acceleration of the block is −5.50î + 2.25k m/s2.
Express your answer in vector form.
Q = N

(c) Determine the force Q when the acceleration of the block is 5.50î − 2.25k m/s2.
Express your answer in vector form.
Q = N

(a) Ah, the block is moving with constant velocity, which means the net force on it is equal to zero. Since there is no friction, we only have the forces P and Q acting on the block. So in order for the net force to be zero, the force Q must be equal in magnitude and opposite in direction to force P. Therefore, Q = -P = (-1.10î - 4.25k) N.

(b) When the acceleration of the block is -5.50î + 2.25k m/s^2, we can use Newton's second law, F = ma, to solve for the force Q. Since the mass (m) of the block is 4.50 kg, we can plug in the values:

Q = m*a = (4.50 kg)*(-5.50î + 2.25k m/s^2) = -24.75î + 10.125k N.

(c) Finally, when the acceleration of the block is 5.50î - 2.25k m/s^2, we can again use Newton's second law to determine the force Q:

Q = m*a = (4.50 kg)*(5.50î - 2.25k m/s^2) = 24.75î - 10.125k N.

Remember, these answers are in vector form! Just kidding, they're actually in clown form. 🤡

(a) When the block moves with constant velocity, the net force acting on the block is zero. Since there is no friction, the only two forces acting on the block are P and Q. So, the force Q should be equal in magnitude but opposite in direction to force P in order to achieve equilibrium.

Q = -P

Since P = (1.10î + 4.25k)N, we can express Q as:

Q = -(1.10î + 4.25k)N
Q = (-1.10î - 4.25k)N

Therefore, the force Q when the block moves with constant velocity is -1.10î - 4.25k N.

(b) When the acceleration of the block is -5.50î + 2.25k m/s², we can use Newton's second law of motion to determine the force Q.

Net force (F_net) = mass (m) * acceleration (a)

F_net = m * a
F_net = (4.50 kg) * (-5.50î + 2.25k m/s²)

Since F_net is the vector sum of forces P and Q, we can write:

F_net = P + Q

Substituting the known value of P, we can solve for Q:

P + Q = (4.50 kg) * (-5.50î + 2.25k m/s²)
Q = (4.50 kg) * (-5.50î + 2.25k m/s²) - P

Substituting the value of P = (1.10î + 4.25k)N, we get:

Q = (4.50 kg) * (-5.50î + 2.25k m/s²) - (1.10î + 4.25k)N

Evaluating the expression, we find:

Q = (-24.75î + 10.125k - 1.10î - 4.25k)N
Q = (-25.85î + 5.875k)N

Therefore, the force Q when the acceleration of the block is -5.50î + 2.25k m/s² is -25.85î + 5.875k N.

(c) Similarly, when the acceleration of the block is 5.50î - 2.25k m/s², we can use Newton's second law of motion to determine the force Q.

Net force (F_net) = mass (m) * acceleration (a)

F_net = m * a
F_net = (4.50 kg) * (5.50î - 2.25k m/s²)

Since F_net is the vector sum of forces P and Q, we can write:

F_net = P + Q

Substituting the known value of P, we can solve for Q:

P + Q = (4.50 kg) * (5.50î - 2.25k m/s²)
Q = (4.50 kg) * (5.50î - 2.25k m/s²) - P

Substituting the value of P = (1.10î + 4.25k)N, we get:

Q = (4.50 kg) * (5.50î - 2.25k m/s²) - (1.10î + 4.25k)N

Evaluating the expression, we find:

Q = (24.75î - 10.125k - 1.10î - 4.25k)N
Q = (23.65î - 14.375k)N

Therefore, the force Q when the acceleration of the block is 5.50î - 2.25k m/s² is 23.65î - 14.375k N.

To solve this problem, we need to apply Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration. Since the block is moving along the table with no friction, the only horizontal forces acting on it are P and Q.

(a) When the block moves with constant velocity, the acceleration is zero. Therefore, the net force acting on the block is also zero. This means that the magnitudes of forces P and Q must be equal in order to cancel each other out. So, the force Q is equal in magnitude but opposite in direction to force P:

Q = -P = -(1.10î + 4.25k) N = -1.10î - 4.25k N

(b) When the acceleration of the block is given as -5.50î + 2.25k m/s^2, we can calculate the net force acting on the block using Newton's second law. Since
F_net = ma,
where F_net is the net force, m is the mass of the block, and a is the acceleration of the block.

Now, the net force is equal to the vector sum of forces P and Q:
F_net = P + Q.

Plugging in the given values:
F_net = (1.10î + 4.25k) N + Q = m(-5.50î + 2.25k) m/s^2.

Rearranging the equation to solve for Q:
Q = m(-5.50î + 2.25k) m/s^2 - (1.10î + 4.25k) N,
Q = (-5.50m + 1.10)î + (2.25m + 4.25)k N.

Substituting the mass m = 4.50 kg:
Q = (-5.50(4.50) + 1.10)î + (2.25(4.50) + 4.25)k N,
Q = (-24.75 + 1.10)î + (10.12 + 4.25)k N,
Q = -23.65î + 14.37k N.

Therefore, the force Q when the acceleration of the block is -5.50î + 2.25k m/s^2 is -23.65î + 14.37k N.

(c) To determine the force Q when the acceleration of the block is 5.50î - 2.25k m/s^2, we can use the same approach as in part (b). Following the steps:

Net force, F_net = m(5.50î - 2.25k) m/s^2.

Rearranging the equation to solve for Q:
Q = m(5.50î - 2.25k) m/s^2 - (1.10î + 4.25k) N,
Q = (5.50m - 1.10)î - (2.25m + 4.25)k N.

Substituting the mass m = 4.50 kg:
Q = (5.50(4.50) - 1.10)î - (2.25(4.50) + 4.25)k N,
Q = (24.75 - 1.10)î - (10.12 + 4.25)k N,
Q = 23.65î - 14.37k N.

Therefore, the force Q when the acceleration of the block is 5.50î - 2.25k m/s^2 is 23.65î - 14.37k N.