# AP Physics 1

## Find the Total Work Done on an Object

Two boys pull a block along the floor for a distance of 4.0 m toward the east. One boy pulls with a force F1 ( magnitude 75 N )
and the other pulls with F2 ( also magnitude 75 N). A friction force fr ( magnitude 25 N ) opposes the motion. Find the following:
a) How much work is done on the block by each force?
b) What is the total work done on the block by all three forces?
c) If the block has a mass of 20 kg and has an intial velocity of $5\frac{m}{s}$.
What is the velocity of the block at the end of the 4.0 m?

### Find the components of each Force that is parallel to the displacement

First we want to find the component of Force 1 that is parallel to the displacement $F_{1\parallel}$

$F_{1\parallel} = F_{1}Cos(\theta)$
$F_{1\parallel} = 75Cos(30)$
$F_{1\parallel} = 64.95 N$

Next we want to find the component of Force 1 that is parallel to the displacement $F_{2\parallel}$

$F_{2\parallel} = F_{1}Cos(\theta)$
$F_{2\parallel} = 75Cos(30)$
$F_{2\parallel} = 64.95 N$

Next we want to find the component of friction that is parallel to the displacement $f_{r}$ note that this is negative since it is facing backwards

$f_{r} = - 25 N$

### Now find the work done by each Force

D is the Displacement and is equal to 4 m

Work done by $F_{1\parallel}$

$F_{1\parallel} = 64.95 N$

$$Work_{F1}=DF_{1\parallel}$$ $$Work_{F1}=(4)64.95$$ $$Work_{F1}=260 J$$

Work done by $F_{2\parallel}$

$F_{2\parallel} = 64.95 N$

$$Work_{F2}=DF_{1\parallel}$$ $$Work_{F2}=(4)64.95$$ $$Work_{F2}=260 J$$

Work done by $f_{r}$. Note the work is
negative since the frictional force
is opposing the direction of motion.

$f_{r} = - 25 N$

$$Work_{fr}=Df_{r}Cos(0^{\circ})$$ $$Work_{fr}=(4)(-25)Cos(0^{\circ})$$ $$Work_{fr}=-100 J$$

### Now Find the total Work

$$Work_{Total} = Work_{F1}+Work_{F2}+Work_{fr}$$ $$Work_{Total} = 260 J + 260 J - 100 J$$ $$Work_{Total} = 420 J$$

### Finally we can Find the final Velocity of the Object

#### Given:

$$Mass = 20 kg$$ $$V_{i}= 5 \frac{m}{s}$$ $$Work_{Total} = 420 J$$ $$Work_{Total} = \Delta KE$$

#### Now to Solve for the Final Velocity

$$Work_{Total} = \Delta KE$$ $$Work_{Total} = KE_{f} - KE_{i}$$ $$Work_{Total} = \frac{1}{2}MV_{f}^2 - \frac{1}{2}MV_{i}^2$$ $$420 = \frac{1}{2}20(V_{f}^2) - \frac{1}{2}20(5^2)$$ $$V_{f} = 8.18 \frac{m}{s}$$