Physics -Work probelm

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In summary, a 47.0 kg projectile is launched at an initial speed of 72.0 m/s and an angle of 39.8° above the horizontal. After 7.15 s, it lands on a hillside without air friction. The projectile's kinetic energy at the highest point of its trajectory can be found using an energy approach, and the height of the impact point can be determined by looking at the y-component of the projectile's motion. The total energy just before it hits the hillside can be calculated using conservation of energy.
  • #1
kmotoao
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i need help on this problem. thx

A 47.0 kg projectile is launched with an initial speed of 72.0 m/s and an angle of 39.8° above the horizontal. The projectile lands on a hillside 7.15 s later. Neglect air friction. (a) What is the projectile’s kinetic energy at the highest point of its trajectory? (b) What is the height of the impact point? (c) What is its total energy just before it hits the hillside?
 
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  • #2
For a) Use an energy approach (obiousvly).

[tex]K=\frac{1}{2}mv_x^2+\frac{1}{2}mv_y^2[/tex]
and [itex]E=K+V[/itex].
You know E from the initial conditions.
At the highest point the velocity in the y-direction is zero.

b) I'd look at the y-component only and write the height as function of time. Then find the height of the impact point.

c) Use conservation of energy again.
 
  • #3


(a) To find the kinetic energy at the highest point of the projectile's trajectory, we can use the equation KE = 1/2mv^2, where m is the mass of the projectile and v is its velocity. Plugging in the given values, we get KE = 1/2(47.0 kg)(72.0 m/s)^2 = 152,064 J. Therefore, the projectile's kinetic energy at the highest point is 152,064 J.

(b) To find the height of the impact point, we can use the equation y = y0 + v0yt - 1/2gt^2, where y0 is the initial height, v0y is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time. Since the projectile lands on a hillside, we can assume that the initial height and final height are the same. Also, at the highest point, the vertical component of the velocity is 0. Plugging in the values, we get y = 0 + 0(7.15 s) - 1/2(9.8 m/s^2)(7.15 s)^2 = 247.023 m. Therefore, the height of the impact point is 247.023 m.

(c) The total energy just before the projectile hits the hillside is equal to the sum of its kinetic energy and potential energy. We already calculated the kinetic energy to be 152,064 J. To find the potential energy, we can use the equation PE = mgh, where m is the mass of the projectile, g is the acceleration due to gravity, and h is the height. Plugging in the values, we get PE = (47.0 kg)(9.8 m/s^2)(247.023 m) = 113,709 J. Therefore, the total energy just before the projectile hits the hillside is 152,064 J + 113,709 J = 265,773 J.
 

1. What is a work problem in physics?

A work problem in physics refers to a situation where an object is being moved by a force over a certain distance. It involves calculating the amount of work done on the object, which is the product of the force applied and the distance moved.

2. How do you calculate work in a physics problem?

To calculate work in a physics problem, you need to multiply the force applied to the object by the distance it is moved in the direction of the force. The unit of work is joules (J), which is equal to newtons (N) multiplied by meters (m).

3. What is the relationship between work and energy in a physics problem?

In physics, work and energy are directly related. When work is done on an object, energy is transferred to it. In other words, work is the transfer of energy from one form to another. This relationship is described by the work-energy theorem: the work done on an object equals the change in its kinetic energy.

4. How do you solve a work problem involving multiple forces?

In a work problem involving multiple forces, you need to calculate the work done by each individual force and then add them together to get the total work. It is important to consider the direction of the forces and their components in the direction of motion.

5. Can work be negative in a physics problem?

Yes, work can be negative in a physics problem. This occurs when the force applied to an object is in the opposite direction of the object's motion. In this case, the work done on the object is considered to be negative, indicating that energy is being taken away from the object.

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