Respuesta :
Answer:
(1). 9.35 m/s.
(2).V2 = 3.22 m/s to the left.
Explanation:
So, we are given the following data or parameters or infomation in the question above as;
(1). "open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. "
(2). "A 15 kg package slides down a chute that makes an angle of 27 degrees below the horizontal."
(3). "The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds."
(4). "The package comes to a stop in the cart after 4 seconds."
(a). So, in order to solve this question, we will be making use of the equations below;
(1). Gravitational potential energy = mass × acceleration due to gravity × height.
(2). Kinetic energy = 1/2 × mass × (velocity)^2.
(3). Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
So, starting from equation (1) above;
Gravitational potential energy = mass × acceleration due to gravity × height.
Gravitational potential energy = 15 × 9.8 × 4 = 588 J.
Kinetic energy = 1/2 × mass × (velocity)^2. = 1/2 × 15 × (3)^2.= 67.5 J.
kinetic energy (final) = 1/2 × 15 × v^2. = 7.5v^2.
Hence, Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
[potential energy (final) = 0].
67.5 + 588 = 7.5v^2.
v = 9.35 m/s
(b). -3 cos (27°) = -2.67 m/s.
15 × -2.67 + 50 × 5 =( 15 + 50) × v2.
=> 209.5 = 65v2.
=> V2 = 3.22 m/s to the left.
(1) The speed of the package before it lands on the cart is 9.35 m/s.
(2) The final speed of the cart is 3.22 m/s
Conservation of energy and momentum:
The gravitational potential energy of a system is given by
PE = mgh
where m is the mass
g is the acceleration due to gravity
and h is the height.
The kinetic energy is given by:
KE = (1/2) mv²
According to the law of conservation of energy:
Kinetic energy (initial) + potential energy (initial) = kinetic energy (final) + potential energy (final).
The final potential energy will be zero since taking the cart as the reference as ground.
PE(initial) = 15 × 9.8 × 4 = 588 J
KE(initial) = (1/2) mv² = 1/2 × 15 × 3² = 67.5 J
KE(final) = 1/2 × 15 × v² = 7.5v²
So,
67.5 + 588 = 7.5v²
v = 9.35 m/s
(b) The package leave at an angle of 27° down the horizontal with velocity 3 m/s, so the horizontal speed is:
v ' = -3 cos (27°) = -2.67 m/s.
Negative sign, as the package is moving opposite to the cart
After the package lands on the cart, both move together, so from conservation of momentum:
15 × (-2.67) + 50 × 5 =( 15 + 50) × V
where V is the speed of cart and package together
209.5 = 65 V
V = 3.22 m/s to the left.
The question was incomplete. After searching it, it is likely that it was as given below:
In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore friction between the cart and the floor. A 15.0-kg package slides down a chute that is inclined at 27∘ from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what is (1) the speed of the package just before it lands in the cart and (2) the final speed of the cart?
Learn more about the conservation of energy:
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