The A box is being pulled by a rope across a level floor as shown in

the diagram. The box is accelerating at 1 m/s2, the coefficient of
kinetic friction is 0.5. What is the tension in the rope?

The diagram is important. Is the rope horizontal? If it is, then

tension-friction=mass*acceleration
tension-mass*g*.5=mass*1
tension=mass(4.9+1)
So you need the mass....

To find the tension in the rope, we first need to calculate the force of kinetic friction acting on the box.

The force of kinetic friction can be calculated using the equation:

Fk = μk * N

where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.

In this case, the box is being pulled across a level floor, so the normal force is equal to the weight of the box, which can be calculated using the equation:

N = m * g

where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Next, we can calculate the force of kinetic friction:

Fk = 0.5 * (m * g)

Since the box is accelerating, there must be a net force acting on the box. The net force can be calculated using the equation:

Fnet = m * a

where Fnet is the net force, m is the mass of the box, and a is the acceleration.

In this case, the net force is equal to the sum of the force applied by the tension in the rope and the force of kinetic friction:

Fnet = T - Fk

Setting Fnet equal to m * a and substituting the expressions for Fk and N, we can solve for the tension T:

m * a = T - (0.5 * m * g)

Simplifying the equation, we find:

T = m * (a + 0.5 * g)

Now that we have the equation for the tension in the rope, we can substitute the given values for a (1 m/s^2) and the coefficient of kinetic friction (0.5), and calculate the tension using the mass of the box.