2 Energy-related equations: Lost and getting started please

[Snomann92]

A worker pushed a 27 kg block 7.4 m along a level floor at constant speed with a force directed 28° below the horizontal.

(a) If the coefficient of kinetic friction is 0.20, how much work was done by the worker's force?

Magnitude of the x component of the applied force = - Magnitude of the kinetic frictional force, but part of the applied force is in the normal force?

F_A,x+mu_k*|F_N|=F_A,y+m*g+F_N?

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Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 2.00 m from the edge of the table, see the figure. Bobby compresses the spring 1.10 cm, but the center of the marble falls 22.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?

No mass, no height, no value for k... I don't even know where to start. :S

 

[anathema]

Hello there,

I would like to address both of your questions.

For the first question about the worker pushing a block, yes, you are correct that the magnitude of the x component of the applied force is equal to the magnitude of the kinetic frictional force. However, in this case, the applied force is not only in the x-direction, but also has a component in the y-direction (FA,y). So the equation would be:

FA,x + FA,y + μk|FN| = mg

Where μk is the coefficient of kinetic friction, FN is the normal force, and mg is the weight of the block.

To solve for the work done by the worker's force, we can use the formula:

W = Fcosθ * d

Where F is the applied force, θ is the angle between the force and the displacement (in this case, 28°), and d is the displacement (7.4 m). So the work done by the worker's force would be:

W = (FA,x + FA,y)cos28° * 7.4 m

Now, for the second question about the game with the spring-loaded gun, without knowing the mass, height, or value for k, it is impossible to determine the exact amount that Rhoda should compress the spring. However, we can use the concept of conservation of energy to solve for the compression distance.

The total energy of the system (spring and marble) is equal to the potential energy stored in the compressed spring and the kinetic energy of the marble when it is fired. So we can write:

PEinitial = KEfinal

Since PE = 1/2kx^2 and KE = 1/2mv^2, we can set these two equations equal to each other and solve for x (the compression distance):

1/2kx^2 = 1/2mv^2

Where v is the velocity of the marble, which can be found using the given information (distance D and height difference of 22.0 cm).

I hope this helps to clarify things for you. Please let me know if you have any further questions. Keep exploring and learning!