A 4 kg block slides along a frictionless horizontal with a speed of 5.1 m/s. After sliding a distance of 7 m, the block makes a smooth transition to a frictionless ramp inclined at an

angle of 13 degrees� to the horizontal.
The acceleration of gravity is 9.81 m/s2.
How far up the ramp does the block slide
before coming momentarily to rest?
Answer in units of m.

To find the distance the block slides up the ramp before coming momentarily to rest, we need to analyze the forces acting on the block.

1. First, let's find the initial velocity of the block when it reaches the ramp. Since the block slides a distance of 7 m with a speed of 5.1 m/s, we can use the formula v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. The acceleration in this case is 0 m/s² (because the ramp is frictionless), so we get:
(5.1 m/s)² = u² + 2(0 m/s²)(7 m)
26.01 m²/s² = u²
Therefore, the initial velocity u is the square root of 26.01 m²/s², which is approximately 5.1 m/s.

2. The block slides up the inclined ramp, experiencing both gravity and the normal force from the ramp. The normal force can be split into two components: one perpendicular to the ramp (N₁) and one parallel to the ramp (N₂). The force of gravity can be split into two components as well: one perpendicular to the ramp (mg₁) and one parallel to the ramp (mg₂).

3. We need to find the acceleration of the block as it moves up the ramp. To do this, we'll first find the net force acting on the block in the direction of motion. We have:
Net force = N₂ - mg₂

4. The acceleration can be found from Newton's second law, which states that the net force is equal to the mass of the object multiplied by its acceleration. So:
N₂ - mg₂ = ma

5. We can express the components of the normal force and the force of gravity as follows:
N₂ = N * sin(13°), where N is the magnitude of the normal force
mg₂ = mg * sin(13°), where mg is the weight of the block (mass * acceleration due to gravity)

6. Substituting the above expressions into the equation in step 4, we get:
N * sin(13°) - mg * sin(13°) = ma

7. Notice that the normal force N (perpendicular to the slope) cancels out the force of gravity mg₁ (also perpendicular to the slope), so we are left only with the parallel components:
N₁ = mg₁ * cos(13°), where N₁ is the magnitude of the normal force perpendicular to the ramp
mg₁ = mg * cos(13°), where mg is the weight of the block (mass * acceleration due to gravity)

8. Since the block is momentarily at rest on the ramp, we have a = 0 m/s². Thus, the equation in step 6 becomes:
N * sin(13°) - mg * sin(13°) = 0

9. We can divide the equation by sin(13°) to solve for N in terms of mg:
N - mg = 0
N = mg

10. Now we can substitute the expression for N in step 7 and find mg in terms of the mass of the block:
mg₁ * cos(13°) = mg
cos(13°) = g₁ / g, where g₁ is the magnitude of the acceleration due to gravity perpendicular to the ramp, and g is the acceleration due to gravity

11. Solving for g₁, we get:
g₁ = g * cos(13°)

12. Now we can substitute the value of g₁ into the equation in step 7 and solve for N₁:
N₁ = mg * cos(13°)

13. We can now substitute the expressions for N₂ and mg₂ into the equation for the net force in step 4, knowing that a = 0 m/s²:
N * sin(13°) - mg * sin(13°) = 0

14. Simplifying the equation, we get:
N = mg

15. Now, let's find the distance up the ramp the block slides before coming momentarily to rest. We can use the equation v² = u² + 2as, where v is the final velocity (which is 0 m/s since the block comes momentarily to rest), u is the initial velocity (5.1 m/s), a is the acceleration (which is unknown), and s is the distance up the ramp. Rearranging the equation, we get:
0 m/s = (5.1 m/s)² + 2a * s

16. Solving for s, we get:
s = -((5.1 m/s)²) / (2a)

17. Since a = g₁ = g * cos(13°) (as found in step 10), we can substitute the value of a into the equation in step 16:
s = -((5.1 m/s)²) / (2 * g * cos(13°))

18. Finally, we can substitute the value of g and solve for s:
s = -((5.1 m/s)²) / (2 * 9.81 m/s² * cos(13°))

Using a calculator, we find that s ≈ -1.029 m.

Therefore, the block slides approximately 1.029 m up the ramp before coming momentarily to rest.

To find the distance up the ramp where the block comes momentarily to rest, we first need to calculate the gravitational force component along the ramp, which will oppose the motion of the block.

1. Calculate the gravitational force component along the ramp:
The gravitational force component acting along the ramp can be found using the formula: F_gravity = m * g * sin(θ), where
- F_gravity is the gravitational force component along the ramp.
- m is the mass of the block (4 kg in this case).
- g is the acceleration due to gravity (9.81 m/s^2 in this case).
- θ is the angle of the ramp (13 degrees).

Plugging in the values:
F_gravity = 4 kg * 9.81 m/s^2 * sin(13 degrees)

2. Calculate the work done against the gravitational force:
The work done against the gravitational force is equal to the change in the block's kinetic energy. This can be calculated using the formula: W_gravity = ΔKE = 0.5 * m * (v_f^2 - v_i^2), where
- W_gravity is the work done against the gravitational force.
- ΔKE is the change in kinetic energy.
- m is the mass of the block (4 kg in this case).
- v_f is the final velocity of the block (0 m/s since it comes to a rest).
- v_i is the initial velocity of the block (5.1 m/s in this case).

Plugging in the values:
W_gravity = 0.5 * 4 kg * (0^2 - 5.1^2) m^2/s^2

3. Set the work done against the gravitational force equal to the gravitational force component along the ramp to find the distance:
Since the work done against the gravitational force is equal to the gravitational force component multiplied by the distance, we can write the equation: W_gravity = F_gravity * d, where
- d is the distance up the ramp where the block comes to rest.

Plugging in the values:
0.5 * 4 kg * (0^2 - 5.1^2) m^2/s^2 = 4 kg * 9.81 m/s^2 * sin(13 degrees) * d

Now, we can solve for d:

0.5 * (0 - 5.1^2) = 9.81 * sin(13 degrees) * d

-13.02 = 9.81 * 0.22495105 * d

-13.02 = 2.20415276 * d

d = -13.02 / 2.20415276

d ≈ -5.903 m

Taking the absolute value, the distance up the ramp where the block comes momentarily to rest is approximately 5.903 meters.

Wt. of block=m*g = 4kg * 9.8N/kg=39.2 N.

Et = Ek + Ep = 0.5m*V^2
Ek + Ep = 0.5*4*(5.1)^2 = 52 J.
0 + Ep = 52 @ hmax.
Ep = 52 @ hmax
Ep = mg*h = 52
39.2h = 52
h = 1.33 m.

Dist. up ramp = h/sinA = 1.33/sin13=
5.91 m.