How Does Friction Impact Work and Energy in Physics Problems?

[7C0A0A5]

In the image {08_53.gif}, a block is moved down an incline a distance of 5.0 m from point A to point B by a force F that is parallel to the incline and has a magnitude of 2.3 N. The magnitude of the frictional force acting on the block is 10 N. If the kinetic energy of the block increases by 35 J between A and B, how much work is done on the block by its weight as the block moves from A to B? In Joules

--I'm not sure how to go about this problem...it looks like the masses drop out somewhere. But I don't understand how to get the answer. Also..is this looking for the work done by friction?


The image {08_35.gif} shows an 8.00 kg stone resting on a spring. The spring is compressed 7.0 cm by the stone.

(a) What is the spring constant?
The answer to this question is 11.211428571428571428571428571429 N/cm
(Sorry, got carried away with the digits )

(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release? In Joules

(c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height? In Joules

(d) What is that maximum height, measured from the release point? In meters


--So if I find (b), then (c) would be mgh, where h = the height to equilibrium right?
--and then (c) would be...duh...'h' plus 30 cm
--So if I'm understanding both (c) and (d) correctly the only thing stopping me is (b). So if you could give me a "huge" shove inthe right direction . I'd be obliged.

 

[Nate810]

For (b), the elastic potential energy of the compressed spring just before the release is equal to the work done in compressing it. The work done is equal to the force applied times the distance the spring is compressed. Thus, the elastic potential energy is equal to F*x = (11.211428571428571428571428571429 N/cm)*(0.07 m) = 0.78 J. For (c), the change in gravitational potential energy is equal to the mass of the stone times the acceleration due to gravity times the change in height. Thus, the change in gravitational potential energy is equal to mgh = (8.00 kg)*(9.8 m/s^2)*(0.30 m) = 235.2 J. For (d), the maximum height of the stone is equal to the initial compression of the spring plus the additional compression of the spring. Thus, the maximum height is equal to 0.07 m + 0.30 m = 0.37 m.

 

[wais]

(a) To find the spring constant, we can use the formula F = kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring. In this case, we know that the spring is compressed by 7.0 cm, and the force applied is the weight of the stone, which is mg = 8.00 kg * 9.8 m/s^2 = 78.4 N. So we have 78.4 N = k * 0.07 m. Solving for k, we get k = 78.4 N / 0.07 m = 1120 N/m. However, the unit for k should be N/cm, so we need to convert 1120 N/m to N/cm by dividing by 100, giving us 11.2 N/cm.

(b) The elastic potential energy of a spring is given by the formula U = 1/2 * k * x^2, where k is the spring constant and x is the displacement of the spring. In this case, the spring is compressed by 30.0 cm, so x = 0.30 m. Plugging in the values, we get U = 1/2 * 11.2 N/cm * (0.30 m)^2 = 0.5 J.

(c) The change in gravitational potential energy is given by the formula ΔU = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the change in height. In this case, the stone has a mass of 8.00 kg and the change in height is from the release point to its maximum height, which we will call h. So ΔU = 8.00 kg * 9.8 m/s^2 * h. However, the question is asking for the change in gravitational potential energy, so we need to subtract the initial potential energy at the release point, which we found in part (b) to be 0.5 J. So the final answer is ΔU = 8.00 kg * 9.8 m/s^2 * h - 0.5 J.

(d) To find the maximum height, we can use the conservation of energy principle, which states that the total energy at any point in time must be equal to the total energy at any other point in