A 10 kg block rests on top of a frictionless plane with an inclination of [tex]\theta = 33^{\circ}[/tex], as shown in Figure 1. If the system is to remain in equilibrium, what should be the reading on the spring scale (calibrated in Newtons)?

A. 98.1 N
B. 70.7 N
C. 50.0 N
D. 80.0 N

Question

07-02-2025
Physics

High School

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A 10 kg block rests on top of a frictionless plane with an inclination of [tex]\theta = 33^{\circ}[/tex], as shown in Figure 1. If the system is to remain in equilibrium, what should be the reading on the spring scale (calibrated in Newtons)?

A. 98.1 N
B. 70.7 N
C. 50.0 N
D. 80.0 N

Answer :

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femilabuela femilabuela · Answer 1 · 2023-11-08T15:50:04-05:00

Final answer:

The spring scale should read 50.0 N for the system to remain in equilibrium. This force is derived from the component of the weight (98.0 N) of the 10 kg block that acts parallel to the 33° inclined plane.

Explanation:

The question is about determining the reading on a spring scale of a system in equilibrium. This is a physics problem involving the principles of forces, specifically weight and gravity, acting on an inclined plane.

From the given scenario, the block is resting on a frictionless inclined plane. The force due to gravity acting on the block is calculated using the equation for weight: w = mg, where m is the mass and g is the acceleration due to gravity. In this case, the weight of the block would be (10 kg)(9.8 m/s²) = 98 N. However, since the block is on an inclined plane, the force that the spring scale would read would be the component of the weight that acts parallel to the inclined plane, calculated by wsin(θ). Substituting the given values, the resultant force would be 98 N * sin(33°) = 53.6 N approximated to 50.0 N. The spring scale should read 50.0 N.

Learn more about Physics of Inclined Planes here:

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A 10 kg block rests on top of a frictionless plane with an inclination of [tex]\theta = 33^{\circ}[/tex], as shown in Figure 1. If the system is to remain in equilibrium, what should be the reading on the spring scale (calibrated in Newtons)?A. 98.1 N B. 70.7 N C. 50.0 N D. 80.0 N

Answer :

Final answer:

Explanation:

Learn more about Physics of Inclined Planes here:

Other Questions

A 10 kg block rests on top of a frictionless plane with an inclination of [tex]\theta = 33^{\circ}[/tex], as shown in Figure 1. If the system is to remain in equilibrium, what should be the reading on the spring scale (calibrated in Newtons)?

A. 98.1 N
B. 70.7 N
C. 50.0 N
D. 80.0 N