Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:


Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:


Where:
d: is the distance traveled = 46.1 m
: is the final speed of the truck = 0 (it stops)
: is the initial speed of the truck = 17.9 m/s
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
Answer:
-2.83 m/s²
Explanation:
- Initial velocity (u) = 34 m/s
- Final velocity (v) = 17 m/s
- Time taken (t) = 6 seconds
❖ Acceleration is defined as the rate of change in velocity with time.
→ a = (v - u)/t
- v denotes final velocity
- a denotes acceleration
- u denotes initial velocity
- t denotes time
→ a = (17 - 34)/6 m/s²
→ a = -17/6 m/s²
<h3>→ Acceleration = -2.83 m/s²</h3>
(Minus sign implies that the velocity is decreasing.)
For simplicity, let's call vector B-A vector C Then C is
Cx = (-6.1 - 2.2)
Cy = (-2.2 - (-6.9)) Or,
Cx = -8.3 Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J