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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?

Sagot :

jacob193

Answer:

Approximately [tex]3.69 \times 10^{3}\; \rm J[/tex].

Explanation:

Let [tex]m[/tex] denote the mass of this cart of bricks and let [tex]g[/tex] denote the gravitational acceleration. Let [tex]\mu[/tex] 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: [tex]W = m \cdot g[/tex].
  • Normal force from the ground: [tex]N[/tex].
  • Friction between the cart and the ground: [tex]f = \mu \cdot N[/tex].
  • Force exerted through the rope, [tex]F[/tex].

Consider the force exerted through the rope in two components:

  • Horizontal component of [tex]F[/tex]: [tex]F \cos(27.3^{\circ})[/tex].
  • Vertical component of [tex]F[/tex]: [tex]F \sin(27.3^{\circ})[/tex].

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

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

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

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

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

[tex]N + F\, \sin(27.3^{\circ}) = m \cdot g[/tex].

[tex]F\, \cos(27.3^{\circ}) = \mu \, N[/tex].

Solve this system of equations for [tex]F[/tex], the size of the force that the rope exerted on the cart or bricks.

Rewrite the first equation to find an expression for [tex]N[/tex]:

[tex]N = m\cdot g - F\, \sin(27.3^{\circ})[/tex].

Substitute this expression into the second equation:

[tex]F\, \cos(27.3^{\circ}) = \mu\, (m\cdot g - F\, \sin(27.3))[/tex].

Rearrange and solve for [tex]F[/tex]:

[tex]\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}[/tex].

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

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

Given that [tex]F \approx 179.67\; \rm N[/tex], the horizontal component of this force would be:

[tex]F\, \cos(27.3^{\circ}) \approx 159.66\; \rm N[/tex].

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

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

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