Question

A 28 kg block is pushed 10.3 m along a level floor at constant speed with an applied force F directed 31 degrees below the horizontal (meaning it is directed somewhat downward). If the coefficient of kinetic friction between the block and the floor is 0.22, what is the magnitude of the applied force F? And by how much does the thermal energy of the block-floor system increase?

Answer 1

(a) 81.2 N

Let's write the equation of the forces along the two directions:

- Horizontal direction:

`F cos \theta -\mu R = 0` (1)

where

F is the applied force

`\theta=31^(\circ)` is the angle at which the force is applied, below the horizontal

`\mu R` is the force of friction, and

`\mu = 0.22` is the coefficient of friction

R is the normal reaction

- Vertical direction:

`R-Fsin \theta - mg = 0` (2)

where

(mg) is the weight of the object, with

m = 28 kg being the mass

g = 9.8 m/s^2 is the acceleration of gravity

We start from eq.(1) to find an expression for R:

`F cos \theta = \mu R\\R=(F cos \theta)/(\mu)` (3)

And substituting into (2), we can find the magnitude of F:

`(F cos \theta)/(\mu)-Fsin \theta - mg = 0\\F((cos \theta)/(\mu)-sin \theta)=mg\\F=(mg)/((cos \theta)/(\mu)-sin \theta)=((28)(9.8))/((cos 31^(\circ))/(0.22)-sin 31^(\circ))=81.2 N`

(b) 717 J

The thermal energy of the block-floor system increases by an amount equal to the work done by the force of friction, therefore:

`\Delta E = W = F_f d= \mu R d` (4)

where

d = 10.3 m is the distance across which the block is moved.

We still have to find the reaction force. We can do it using eq.(3):

`R=(F cos \theta)/(\mu)=((81.2)(cos 31))/(0.22)=316.4 N`

So now we can use eq.(4) to find the work done, and the increase in thermal energy:

`\Delta E=(0.22)(316.4)(10.3)=717 J`