A cart loaded with bricks has a total mass of 13.5 kg and is pulled at constant speed by a rope.The rope is inclined at 27.9 degrees above the horizontal and the cart moves 13mts on a horizontal floor.The co efficient of kinetic friction between ground and cart is 0.3.The acceleration of gravity is 9.8 m/s^2.How much work is done on the cart by the rope.
Work done by rope = W
= (rope force)*cos 27.9)*13 meters
= (friction force)*13 m
The rope force F is given by the equation
horizontal component of F
= friction force
(since there is zero acceleration)
F*cos27.9 - (M*g-Fsin27.9)*0.30 = 0
0.8837 F = (132.3 - 0.4679 F)0.30
(0.8837 F + 0.1404 F)= 39.69
F = ?
W = ?
To find the work done on the cart by the rope, we need to calculate the force applied by the rope and the distance over which the force is applied.
First, let's calculate the force applied by the rope. Since the cart is moving at a constant speed, the force applied by the rope must be equal in magnitude and opposite in direction to the force of kinetic friction acting on the cart.
The force of kinetic friction can be calculated using the equation:
\(f_{\text{friction}} = \mu \times f_{\text{norm}}\)
where:
\(f_{\text{friction}}\) is the force of friction,
\(\mu\) is the coefficient of kinetic friction,
\(f_{\text{norm}}\) is the normal force.
The normal force can be calculated using the equation:
\(f_{\text{norm}} = m \times g \times \cos(\theta)\)
where:
\(m\) is the mass of the cart,
\(g\) is the acceleration due to gravity,
\(\theta\) is the angle of inclination of the rope.
Plugging in the given values, we have:
\(m = 13.5 \, \text{kg}\)
\(\mu = 0.3\)
\(g = 9.8 \, \text{m/s}^2\)
\(\theta = 27.9^\circ\)
\(f_{\text{norm}} = 13.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(27.9^\circ)\)
Now, we can calculate \(f_{\text{friction}}\) using:
\(f_{\text{friction}} = 0.3 \times f_{\text{norm}}\)
Next, let's calculate the distance over which the force is applied. In this case, it is given that the cart moves 13 m on a horizontal floor.
Finally, we can calculate the work done by the rope using the equation:
\(W = f \times d \times \cos(\theta)\)
where:
\(W\) is the work done,
\(f\) is the force applied,
\(d\) is the distance over which the force is applied,
\(\theta\) is the angle between the force and the direction of motion.
Plugging in the values we calculated, we have:
\(W = f_{\text{friction}} \times 13 \, \text{m} \times \cos(0^\circ)\)
Since the cart is moving horizontally, the angle between the force and the direction of motion is 0 degrees.
Simplifying this equation, we get:
\(W = f_{\text{friction}} \times 13 \, \text{m}\)
Now, all you need to do is plug in the values for \(f_{\text{friction}}\) and calculate the final value of work done.
To find the work done on the cart by the rope, we need to first determine the force applied by the rope and the distance over which the force is applied.
Given:
Mass of the cart (m) = 13.5 kg
Angle of inclination (θ) = 27.9 degrees
Distance traveled horizontally (d) = 13 m
Coefficient of kinetic friction (μ) = 0.3
Acceleration due to gravity (g) = 9.8 m/s^2
First, let's find the force of gravity acting on the cart (F_gravity):
F_gravity = m * g
F_gravity = 13.5 kg * 9.8 m/s^2
F_gravity = 132.3 N
Next, let's find the force of friction (F_friction):
F_friction = μ * (normal force)
The normal force (F_normal) is the force exerted by the ground on the cart and is equal to the force of gravity acting vertically downwards.
F_normal = F_gravity = 132.3 N
F_friction = μ * F_normal
F_friction = 0.3 * 132.3 N
F_friction = 39.69 N
Since the cart is pulled at constant speed, the net force acting on it is zero. This means that the force applied by the rope is equal in magnitude and opposite in direction to the force of friction.
Force applied by the rope (F_rope) = F_friction = 39.69 N
Now, we can find the work done on the cart by the rope:
Work = force * distance * cos(θ)
Work = F_rope * d * cos(θ)
Substituting the values:
Work = 39.69 N * 13 m * cos(27.9 degrees)
To calculate the work, convert the angle from degrees to radians:
27.9 degrees * (π/180) = 0.487 radians
Work = 39.69 N * 13 m * cos(0.487 radians)
Now, use the cosine function to calculate the work:
cos(0.487 radians) ≈ 0.8945
Work ≈ 39.69 N * 13 m * 0.8945
Work ≈ 545.098 J
Therefore, the work done on the cart by the rope is approximately 545.098 Joules.