Example Question - calculating acceleration
Here are examples of questions we've helped users solve.
Calculating Acceleration With Friction
The net force on the block is the difference between the applied force and the frictional force. Net force = Applied force - Frictional force Net force = 25 N - 15 N Net force = 10 N To find the acceleration (a), use Newton's second law, \( F = ma \), where F is the net force and m is the mass. Solving for acceleration \( a \), \( a = \frac{F}{m} \) Substitute the known values into the equation, \( a = \frac{10 \, \text{N}}{5 \, \text{kg}} \) The acceleration of the block is \( 2 \, \text{m/s}^2 \).
Calculating Acceleration of an Object
The image you've provided includes a question which states: "Solve the following word problem. 1. If a 4 kg object experiences a net force of 12 N, what will be the object's acceleration?" To solve this problem, we'll use Newton's Second Law of Motion, which is defined as: F = ma where F is the net force applied to the object, m is the mass of the object, and a is the acceleration of the object. We are given: F (net force) = 12 N m (mass) = 4 kg We need to find the acceleration (a). Firstly, we rearrange the equation to solve for acceleration: \( a = \frac{F}{m} \) Now we can substitute in the values we have: \( a = \frac{12 \text{ N}}{4 \text{ kg}} \) Now doing the calculation: \( a = 3 \text{ m/s}^2 \) The object's acceleration would be 3 meters per second squared.
Calculating Acceleration Due to Gravity from Velocity and Time
The photo shows three questions, of which the first one (#4) is: If a ball is dropped and attains a velocity of 29.4 m/s in 3.00 s, what is the acceleration due to gravity? To solve this, you can use the following kinematic equation for uniformly accelerated motion: \[ v = u + at \] where: - \( v \) is the final velocity - \( u \) is the initial velocity (which is 0 m/s, since the ball is dropped) - \( a \) is the acceleration (which we are trying to find, and we expect it to be the acceleration due to gravity) - \( t \) is the time Given that the initial velocity \( u = 0 \) m/s, the final velocity \( v = 29.4 \) m/s, and the time \( t = 3.00 \) s, you can plug these values into the equation to find \( a \): \[ 29.4 = 0 + a \times 3.00 \] Now solve for \( a \): \[ a = \frac{29.4}{3.00} \] \[ a \approx 9.8 \, \text{m/s}^2 \] Thus, the acceleration due to gravity, \( g \), is approximately 9.8 m/s², which is in fact the standard value for gravitational acceleration near the Earth's surface.
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