A 4 kg block slides down a frictionless incline making an angle of 60° with the horizontal.
(a) What is the total work done on the block when the block slides 2 m (measured along the incline)?
(b) What is the speed of the block after it has slid 1.5 m if it starts from rest?
(c) What is its speed after 1.5 m if it starts with an initial speed of 2 m/s?
To solve these problems, we can use the principles of work and energy.
(a) The total work done on an object is equal to the change in its kinetic energy. In this case, since the block starts from rest, its initial kinetic energy is zero. Therefore, the total work done will be equal to the final kinetic energy.
To find the final velocity, we can use the following equation:
m*g*h = 1/2*m*v^2
where m is the mass of the block (4 kg), g is the acceleration due to gravity (9.8 m/s^2), h is the height the block slides along the incline (2 m), and v is the final velocity.
First, let's calculate the vertical height component (h_vert) and the length of the incline (s_incline):
h_vert = h * sin(60°)
= 2 * sin(60°)
= 2 * √3 / 2
= √3 m
s_incline = h * cos(60°)
= 2 * cos(60°)
= 2 * 1/2
= 1 m
Now, we can find the total work done:
W_total = m * g * h_vert
= 4 kg * 9.8 m/s^2 * √3 m
≈ 68 J
Therefore, the total work done on the block when it slides 2 m along the incline is approximately 68 Joules.
(b) To find the final speed of the block after sliding 1.5 m, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the system (block) is equal to the final mechanical energy.
The initial mechanical energy consists of only potential energy:
E_initial = m * g * h_initial
= 4 kg * 9.8 m/s^2 * 1.5 m
≈ 59 J
The final mechanical energy consists of both kinetic and potential energy:
E_final = 1/2 * m * v^2 + m * g * h_final
Since we know that E_initial = E_final, we can set up the equation:
m * g * h_initial = 1/2 * m * v^2 + m * g * h_final
Plugging in the given values:
4 kg * 9.8 m/s^2 * 1.5 m = 1/2 * 4 kg * v^2 + 4 kg * 9.8 m/s^2 * 0
Simplifying the equation:
58.8 J = 2 kg * v^2
Dividing both sides by 2 kg:
v^2 ≈ 29.4 m^2/s^2
Taking the square root:
v ≈ 5.42 m/s
Therefore, the speed of the block after sliding 1.5 m, starting from rest, is approximately 5.42 m/s.
(c) If the block starts with an initial speed of 2 m/s, we can use the principle of conservation of mechanical energy again. The initial mechanical energy of the system (block) is equal to the final mechanical energy.
The initial mechanical energy consists of kinetic and potential energy:
E_initial = 1/2 * m * v_init^2 + m * g * h_initial
The final mechanical energy also consists of kinetic and potential energy:
E_final = 1/2 * m * v_final^2 + m * g * h_final
Since we know that E_initial = E_final, we can set up the equation:
1/2 * m * v_init^2 + m * g * h_initial = 1/2 * m * v_final^2 + m * g * h_final
Plugging in the given values:
1/2 * 4 kg * (2 m/s)^2 + 4 kg * 9.8 m/s^2 * 1.5 m = 1/2 * 4 kg * v_final^2 + 4 kg * 9.8 m/s^2 * 0
Simplifying the equation:
8 J + 58.8 J = 2 kg * v_final^2
v_final^2 ≈ 33.4 m^2/s^2
Taking the square root:
v_final ≈ 5.78 m/s
Therefore, the speed of the block after sliding 1.5 m, starting with an initial speed of 2 m/s, is approximately 5.78 m/s.