Question

A cart loaded with bricks has a total mass of 24.7kg and is pulled at constant speed by a rope. The rope is inclined at 27.3° above the horizontal and the cart moves 23.1m on a horizontal floor. The coefficient of kinetic friction is between ground and cart is 0.8.
The acceleration of gravity is 9.8m/s^2.
How much work is done on the cart by the rope?

Answer 1

Answer:

Approximately $3.69 \times 10^{3}\; \rm J$.

Explanation:

Let $m$ denote the mass of this cart of bricks and let $g$ denote the gravitational acceleration. Let $\mu$ denote the constant of kinetic friction between the ground and this cart of bricks.

Refer to the diagram attached. There are four forces on this cart of bricks:

• Weight of the cart and the bricks: $W = m \cdot g$.
• Normal force from the ground: $N$.
• Friction between the cart and the ground: $f = \mu \cdot N$.
• Force exerted through the rope, $F$.

Consider the force exerted through the rope in two components:

• Horizontal component of $F$: $F \cos(27.3^{\circ})$.
• Vertical component of $F$: $F \sin(27.3^{\circ})$.

Since the velocity of this cart is constant, forces on this cart would be balanced. The following would be equal:

• Vertically, the normal force $N$ and the vertical component of $F$ should balance the weight of the cart and the bricks: $N + F\, \sin(27.3^{\circ}) = W$.
• Horizontally, the horizontal component of $F$ should balance the friction between the cart of bricks and the ground: $F\, \cos(27.3^{\circ}) = f$.

However, $f = \mu \, N$. Thus, the second equation would be equivalent to $F\, \cos(27.3^{\circ}) = \mu \, N$.

The weight of this cart of bricks is $W = m \cdot g$. The first equation would be equivalent to $N + F\, \sin(27.3^{\circ}) = m \cdot g$.

The value of $\text{ m . g , and \mu }$ are all given. Thus, these two equations is a system of two equations for two unknowns, $\text{ N and F }$:

$N + F\, \sin(27.3^{\circ}) = m \cdot g$.

$F\, \cos(27.3^{\circ}) = \mu \, N$.

Solve this system of equations for $F$, the size of the force that the rope exerted on the cart or bricks.

Rewrite the first equation to find an expression for $N$:

$N = m\cdot g - F\, \sin(27.3^{\circ})$.

Substitute this expression into the second equation:

$F\, \cos(27.3^{\circ}) = \mu\, (m\cdot g - F\, \sin(27.3))$.

Rearrange and solve for $F$:

$\begin{aligned}F &= \frac{\mu \cdot m \cdot g}{\cos(27.3^{\circ}) + \sin(27.3^{\circ})} \\ &= \frac{0.8 \times 24.7\; \rm kg \times 9.81\; \rm m\cdot s^{-2}}{\cos(27.3^{\circ}) + \sin(27.3^{\circ})} \\ &\approx 179.67\; \rm N\end{aligned}$.

Thus, the size of the force that the rope exerted on the cart would be approximately $179.67\; \rm N$.

The floor is horizontal. Thus, the vertical displacement of this cart of bricks would be $0$. The vertical component of $F$ would thus have done no work on the cart $0\!$. The entirety of the work that $F\!$ does on this cart of bricks would come from the horizontal component of this force.

Given that $F \approx 179.67\; \rm N$, the horizontal component of this force would be:

$F\, \cos(27.3^{\circ}) \approx 159.66\; \rm N$.

The horizontal displacement of this cart of bricks is $23.1\; \rm m$. Accordingly, the work that the horizontal component of $F$ did on this cart of bricks would be approximately $159.66\; \rm m \times 23.1\; \rm m \approx 3.69 \times 10^{3}\; \rm J$.

Thus, the overall work that this rope did on this cart of bricks would be approximately $3.69 \times 10^{3}\; \rm J$.