/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A block of mass 15.0 kg rests on... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of mass 15.0 kg rests on a horizontal table. A force of \(50.0 \mathrm{N}\) is applied vertically downward on the block. Calculate the force that the block exerts on the table.

Short Answer

Expert verified
The force exerted by the block on the table is 197 N.

Step by step solution

01

- Identify the forces acting on the block

There are two forces acting on the block: the gravitational force (weight) and the applied force. The gravitational force can be calculated using the formula: \(F_g = mg\), where \(m = 15.0 \text{ kg}\) and \(g = 9.8 \text{ m/s}^2\).
02

- Calculate the gravitational force

Use the formula \(F_g = mg\):\[ F_g = 15.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 147 \text{ N} \]
03

- Determine the total downward force

Add the gravitational force and the applied force to find the total downward force: \(F_{total} = F_g + F_{applied} = 147 \text{ N} + 50.0 \text{ N} = 197 \text{ N}\)
04

- Calculate the force exerted by the block on the table

The force that the block exerts on the table is equal to the total downward force: \[ F_{table} = 197 \text{ N} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

gravitational force
Gravitational force is the attractive force exerted by the Earth on objects, pulling them towards its center. Understanding this force is crucial in physics, as it's a constant force that affects everything on our planet. To calculate gravitational force, we use the equation: \( F_g = mg \) where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, roughly \( 9.8 \text{ m/s}^2 \). In our exercise, the block's mass (\( m \)) is 15.0 kg. Plugging the values in, we get: \[ F_g = 15.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 147 \text{ N} \] This 147 N force is the gravitational pull that the Earth exerts on the block. Understanding this concept is key to figuring out other forces acting on the block.
applied force
An applied force is any force that is applied to an object by a person or another object. In the given problem, a 50.0 N force is applied vertically downward on the block. This applied force increases the overall force acting on the block, adding to the force due to gravity. In simpler terms, it's like adding extra weight to the block.

When calculating the total downward force on the block due to both gravity and the applied force, we add these two forces together. This leads us to the next section on total downward force.
total downward force
The total downward force on an object is the sum of all forces acting downward. In our case, it includes both the gravitational force and the applied force. The formula is: \[ F_{total} = F_g + F_{applied} \] For the block:
  • Gravitational force (\( F_g \)) = 147 N
  • Applied force (\( F_{applied} \)) = 50.0 N
Adding these together, we get: \[ F_{total} = 147 \text{ N} + 50.0 \text{ N} = 197 \text{ N} \]

This total downward force (197 N) is the force exerted by the block on the table, ensuring the table supports this full weight. Grasping the concept of total downward force helps you understand how multiple forces interact and add up in the real world, affecting the stability and balance of objects.

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Most popular questions from this chapter

Your brand new convertible Ferrari is parked \(15 \mathrm{m}\) from its garage when it begins to rain. You do not have time to get the keys so you begin to push the car towards the garage. If the maximum acceleration you can give the car is \(2.0 \mathrm{m} \mathrm{s}^{-2}\) by pushing and \(3.0 \mathrm{m} \mathrm{s}^{-2} \mathrm{by}\) pulling back on the car, find the least time it takes to put the car in the garage. (Assume that the car, as well as the garage, are point objects.)

A rocket accelerates vertically upwards from rest with a constant acceleration of \(4.00 \mathrm{m} \mathrm{s}^{-2} .\) The fuel lasts for \(5.00 \mathrm{s}\) (a) What is the maximum height achieved by this rocket? (b) When does the rocket reach the ground again? (c) Sketch a graph to show the variation of the velocity of the rocket with time from the time of launch to the time it falls to the ground. (Take the acceleration due to gravity to be \(10.0 \mathrm{m} \mathrm{s}^{-2} .\)

Show that an alternative formula for kinetic energy is \(E_{\mathrm{k}}=\frac{\rho^{2}}{2 m},\) where \(p\) is the momentum of the mass \(m\). This is very useful when dealing with collisions.

An elevator starts on the ground floor and stops on the 10 th floor of a high- rise building. The elevator picks up a constant speed by the time it reaches the 1 st floor and decelerates to rest between the 9 th and 10th floors. Describe the energy transformations taking place between the 1 st and 9th floors.

A ball of mass \(m\) is dropped from a height of \(h_{1}\) and rebounds to a height of \(h_{2} .\) The ball is in contact with the floor for a time interval of \(\tau\). (a) Show that the average net force on the ball is given by $$F=m \frac{\sqrt{2 g h_{1}}+\sqrt{2 g h_{2}}}{\tau}$$ (b) If \(h_{1}=8.0 \mathrm{m}, h_{2}=6.0 \mathrm{m}, \tau=0.125 \mathrm{s}\), \(m=0.250 \mathrm{kg},\) calculate the average force exerted by the ball on the floor.

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