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How much force must be applied to make a 4.75 kg block just start to move up an incline set at 60 , the coefficient of friction is \( \square \mathrm{s}=0.23 \) ? \[ 2<2 / 2 \]

Ask by Barber Tran. in the United States
Feb 18,2025

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A force of approximately 57.33 N must be applied to make the 4.75 kg block just start to move up the 60-degree incline with a coefficient of friction of 0.23.

Solución

To find the force required to make a 4.75 kg block just start to move up an incline set at 60 degrees, we need to consider the forces acting on the block. The forces acting on the block are: 1. Weight of the block (W) 2. Normal force (N) exerted by the incline 3. Frictional force (F) opposing the motion 4. Applied force (F_app) to make the block move up the incline The weight of the block (W) can be calculated using the formula: \[ W = m \times g \] where: - \( m \) is the mass of the block (4.75 kg) - \( g \) is the acceleration due to gravity (approximately 9.81 m/s^2) The normal force (N) is equal to the weight of the block since the block is at rest initially. The frictional force (F) can be calculated using the formula: \[ F = \mu \times N \] where: - \( \mu \) is the coefficient of friction (0.23) The applied force (F_app) is the force required to make the block move up the incline. This force is equal to the sum of the weight of the block and the frictional force. Let's calculate the weight of the block: \[ W = 4.75 \, \text{kg} \times 9.81 \, \text{m/s}^2 \] Now, let's calculate the frictional force: \[ F = 0.23 \times W \] Finally, we can calculate the applied force (F_app) to make the block move up the incline: \[ F_{app} = W + F \] Let's calculate these values step by step. Calculate the value by following steps: - step0: Calculate: \(4.75\times 9.81\) - step1: Multiply the numbers: \(46.5975\) The weight of the block is approximately 46.5975 N. Now, let's calculate the frictional force: \[ F = 0.23 \times W \] \[ F = 0.23 \times 46.5975 \] \[ F = 10.737 \, \text{N} \] Finally, the applied force (F_app) to make the block move up the incline is: \[ F_{app} = W + F \] \[ F_{app} = 46.5975 + 10.737 \] \[ F_{app} = 57.3345 \, \text{N} \] Therefore, the force required to make a 4.75 kg block just start to move up an incline set at 60 degrees is approximately 57.3345 N.

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To calculate the force required to start moving the block up the incline, we first need to consider both the gravitational force acting down the slope and the frictional force opposing the motion. The force due to gravity can be calculated as \( mg \sin(\theta) \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (~9.81 m/s²), and \( \theta \) is the angle of the incline (60°). The normal force is \( N = mg \cos(\theta) \), and the force of friction is given by \( f_f = \mu N \), where \( \mu \) is the coefficient of friction (0.23). We can thus sum up these forces to find the total force required to get the block moving: 1. Calculate gravitational force down the incline: \( F_g = 4.75 \, \text{kg} \times 9.81 \, \text{m/s}² \times \sin(60°) \). 2. Calculate normal force: \( N = 4.75 \, \text{kg} \times 9.81 \, \text{m/s}² \times \cos(60°) \). 3. Calculate frictional force: \( f_f = 0.23 \times N \). 4. Total force required to overcome both gravity and friction: \( F = F_g + f_f \). Plug in the numbers, and you should find that you need to apply a force of approximately 38.63 N to get the block just starting to move up the incline!

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