Q 2 **Centripetal Force and Tension Analysis on a Frictionless Table**A 500 g block on top of a frictionless table top is being spun around in a circle by a string that is angled upwards at 25°. If the block is moving at a constant 3 m/s in a circle with an 85 cm radius, determine the normal force on the block and the tension in the string.**Diagram Overview:**The diagram illustrates a block of mass 500 grams placed on a frictionless table. The block is connected to a string inclined at an angle of 25° from the horizontal axis. The block is spinning at a constant speed of 3 m/s in a circular path with a radius of 85 cm (0.85 m).**Given:**- Mass of the block (m): 500 g (0.5 kg)- Speed of the block (v): 3 m/s- Radius of the circle (r): 85 cm (0.85 m)- Angle of the string with horizontal (θ): 25°**Find:**1. Normal force (N) on the block.2. Tension (T) in the string.**Explanation:**1. **Centripetal Force Calculation:** The centripetal force \( F_c \) required to keep the block moving in a circular path is given by: \[ F_c = \frac{mv^2}{r} \] Substituting the values: \[ F_c = \frac{(0.5 \, \text{kg}) (3 \, \text{m/s})^2}{0.85 \, \text{m}} = \frac{4.5}{0.85} \approx 5.29 \, \text{N} \]2. **Vertical Forces Analysis:** The vertical forces acting on the block are the gravitational force \( mg \) (downward) and the vertical component of the tension \( T \sin \theta \) (upward). For the block to remain in equilibrium vertically: \[ T \sin \theta = mg \] The gravitational force is: \[ mg = 0.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 4.9 \

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A 500 g block on top of a frictionless table top is being spun around in a circle by a string that is angled upwards at 25°. If the block is moving at a constant 3 m/s in a circle with an 85 cm radius, determine the normal force on the block and the tension in the string. 85 сm 25°

A 500 g block on top of a frictionless table top is being spun around in a circle by a string that is angled upwards at 25°. If the block is moving at a constant 3 m/s in a circle with an 85 cm radius, determine the normal force on the block and the tension in the string. 85 сm 25°

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

Question

Q 2

**Centripetal Force and Tension Analysis on a Frictionless Table**

A 500 g block on top of a frictionless table top is being spun around in a circle by a string that is angled upwards at 25°. If the block is moving at a constant 3 m/s in a circle with an 85 cm radius, determine the normal force on the block and the tension in the string.

**Diagram Overview:**

The diagram illustrates a block of mass 500 grams placed on a frictionless table. The block is connected to a string inclined at an angle of 25° from the horizontal axis. The block is spinning at a constant speed of 3 m/s in a circular path with a radius of 85 cm (0.85 m).

**Given:**
- Mass of the block (m): 500 g (0.5 kg)
- Speed of the block (v): 3 m/s
- Radius of the circle (r): 85 cm (0.85 m)
- Angle of the string with horizontal (θ): 25°

**Find:**
1. Normal force (N) on the block.
2. Tension (T) in the string.

**Explanation:**

1. **Centripetal Force Calculation:**
The centripetal force \( F_c \) required to keep the block moving in a circular path is given by:
\[
F_c = \frac{mv^2}{r}
\]
Substituting the values:
\[
F_c = \frac{(0.5 \, \text{kg}) (3 \, \text{m/s})^2}{0.85 \, \text{m}} = \frac{4.5}{0.85} \approx 5.29 \, \text{N}
\]

2. **Vertical Forces Analysis:**
The vertical forces acting on the block are the gravitational force \( mg \) (downward) and the vertical component of the tension \( T \sin \theta \) (upward). For the block to remain in equilibrium vertically:
\[
T \sin \theta = mg
\]

The gravitational force is:
\[
mg = 0.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 4.9 \

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