(a) 81.2 N
Let's write the equation of the forces along the two directions:
- Horizontal direction:
[tex]F cos \theta -\mu R = 0[/tex] (1)
where
F is the applied force
[tex]\theta=31^{\circ}[/tex] is the angle at which the force is applied, below the horizontal
[tex]\mu R[/tex] is the force of friction, and
[tex]\mu = 0.22[/tex] is the coefficient of friction
R is the normal reaction
- Vertical direction:
[tex]R-Fsin \theta - mg = 0[/tex] (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:
[tex]F cos \theta = \mu R\\R=\frac{F cos \theta}{\mu}[/tex] (3)
And substituting into (2), we can find the magnitude of F:
[tex]\frac{F cos \theta}{\mu}-Fsin \theta - mg = 0\\F(\frac{cos \theta}{\mu}-sin \theta)=mg\\F=\frac{mg}{\frac{cos \theta}{\mu}-sin \theta}=\frac{(28)(9.8)}{\frac{cos 31^{\circ}}{0.22}-sin 31^{\circ}}=81.2 N[/tex]
(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:
[tex]\Delta E = W = F_f d= \mu R d[/tex] (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):
[tex]R=\frac{F cos \theta}{\mu}=\frac{(81.2)(cos 31)}{0.22}=316.4 N[/tex]
So now we can use eq.(4) to find the work done, and the increase in thermal energy:
[tex]\Delta E=(0.22)(316.4)(10.3)=717 J[/tex]
