A block of mass m = 5.90 kg is released from rest from point circled A and slides on the frictionless track shown in the figure below. (Assume ha = 7.30 m.)
point b is 3.20 m and point c is 2.00m.
(a) Determine the block's speed at points circled B and circled C.
point circled B m/s
point circled C m/s
(b) Determine the net work done by the gravitational force on the block as it moves from point circled A to point circled C.
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(a) Well, it seems like this block is on the move! Let's calculate its speed at points circled B and circled C.
We can use the conservation of energy to solve this problem. From point A to point B, there is a height change of h = 7.30 m. The potential energy at point A will be converted to kinetic energy at point B.
Let's start by finding the velocity at point B. We can use the equation:
mg(ha - hb) = (1/2)mvb^2,
where m is the mass of the block, g is the acceleration due to gravity, ha is the height at point A, hb is the height at point B, and vb is the velocity at point B.
Plugging in the values, we have:
5.90 kg * 9.8 m/s^2 * (7.30 m - 3.20 m) = (1/2) * 5.90 kg * vb^2.
Simplifying, we get:
202.66 J = 2.95 kg * vb^2.
Solving for vb, we find:
vb = sqrt(202.66 J / 2.95 kg) ≈ 8.13 m/s.
So, the velocity at point B is approximately 8.13 m/s. Now, let's move on to find the velocity at point C.
From point B to point C, there is a height change of hc = 2.00 m. Again, the potential energy at point B will be converted to kinetic energy at point C.
Using the same equation as before, we can find the velocity at point C:
mg(hb - hc) = (1/2)mvc^2.
Plugging in the values, we have:
5.90 kg * 9.8 m/s^2 * (3.20 m - 2.00 m) = (1/2) * 5.90 kg * vc^2.
Simplifying, we get:
34.686 J = 2.95 kg * vc^2.
Solving for vc, we find:
vc = sqrt(34.686 J / 2.95 kg) ≈ 5.82 m/s.
So, the velocity at point C is approximately 5.82 m/s.
(b) Alright, let's move on to part (b) now. We need to determine the net work done by the gravitational force on the block as it moves from point circled A to point circled C.
The net work done by a force is equal to the change in kinetic energy of an object. In this case, the only significant force acting on the block is gravity.
We can use the work-energy theorem to find the net work done by the gravitational force:
Work = ΔKE = KEfinal - KEinitial,
where Work is the net work done, ΔKE is the change in kinetic energy, KEfinal is the final kinetic energy, and KEinitial is the initial kinetic energy.
At point circled A, the block is at rest, so its initial kinetic energy is zero. At point circled C, the block has a kinetic energy of (1/2)mv^2, where v is the velocity at point C.
Using the equation, we have:
Work = (1/2)mv^2 - 0 = (1/2) * 5.90 kg * (5.82 m/s)^2.
Simplifying, we find:
Work ≈ 97.77 J.
So, the net work done by the gravitational force on the block as it moves from point circled A to point circled C is approximately 97.77 J.
Keep rolling with those physics problems!
To solve this problem, we can use the principle of conservation of mechanical energy. The mechanical energy of the system (block) is conserved as there is no friction acting on it. The mechanical energy is given by the sum of kinetic energy (K) and potential energy (U).
(a) To determine the block's speed at points circled B and circled C, we need to find the mechanical energies at both points.
At point circled A, the block is at rest, so its kinetic energy is zero (K=0). The potential energy at point circled A is given by the equation U = m*g*h, where m is the mass of the block, g is the acceleration due to gravity, and h is the height at point A. Plugging in the given values, we have:
U_A = m*g*h_A = (5.90 kg) * (9.8 m/s^2) * (7.30 m) = 319.244 J
At point circled B, the block has only kinetic energy. To find the kinetic energy, we can equate the mechanical energy at point A to the kinetic energy at point B:
U_A = K_B
319.244 J = (1/2) * m * v_B^2
Solving for v_B, we have:
v_B = sqrt((2 * U_A) / m) = sqrt((2 * 319.244 J) / 5.90 kg) ≈ 10.61 m/s
Therefore, the block's speed at point circled B is approximately 10.61 m/s.
At point circled C, the block reaches its highest potential energy, so its kinetic energy is zero (K=0). The potential energy at point circled C is given by the equation U = m*g*h, where m is the mass of the block, g is the acceleration due to gravity, and h is the height at point C (which is zero). Plugging in the given values, we have:
U_C = m*g*h_C = (5.90 kg) * (9.8 m/s^2) * (0) = 0 J
Therefore, the block's speed at point circled C is zero (v_C = 0 m/s).
(b) The net work done by the gravitational force on the block as it moves from point circled A to point circled C can be calculated as the difference in mechanical energy between these two points:
Work = U_C - U_A = 0 J - 319.244 J = -319.244 J
Therefore, the net work done by the gravitational force on the block as it moves from point circled A to point circled C is -319.244 J.
To answer these questions, we need to apply the principles of conservation of energy and work-energy theorem.
(a) To determine the block's speed at points circled B and circled C, we can use the principle of conservation of energy. At point A, the block has gravitational potential energy, which is converted into kinetic energy as it moves down the track.
1. Determine the gravitational potential energy at point A:
The gravitational potential energy (PE) is given by the equation: PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, m is 5.9 kg, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is 7.30 m. Plugging in the values, we get:
PE_A = 5.9 kg * 9.8 m/s^2 * 7.30 m.
2. Determine the total mechanical energy at point A:
The total mechanical energy (E) at point A is the sum of the gravitational potential energy and the kinetic energy. So, E_A = PE_A + KE_A, where KE_A is the initial kinetic energy (which is zero since the block is released from rest).
3. Determine the total mechanical energy at points B and C:
Since there is no friction, the mechanical energy at any point remains constant. So, E_B = E_C = E_A.
4. Determine the speed at points B and C:
The kinetic energy (KE) is given by the equation: KE = 0.5 * m * v^2, where m is the mass and v is the velocity. At points B and C, the kinetic energy is equal to the total mechanical energy (E). So, KE_B = KE_C = E.
Now, plug in the values of E_A into the equation KE_B = KE_C = E and solve for v (the speed at points B and C).
(b) To determine the net work done by the gravitational force on the block as it moves from point A to point C, we can use the work-energy theorem. The work done by a force is equal to the change in kinetic energy.
1. Determine the change in kinetic energy:
The change in kinetic energy (ΔKE) is calculated by subtracting the initial kinetic energy (KE_A) from the final kinetic energy (KE_C). ΔKE = KE_C - KE_A.
2. Determine the net work done:
The net work done (W) is equal to the change in kinetic energy. So, W = ΔKE.
Now, plug in the values of KE_A and KE_C into the equation ΔKE = KE_C - KE_A to find the net work done by the gravitational force on the block.