solution\nfind the mass of the system from the definition of weight, w = mg.\nfind the acceleration of the…

solution\nfind the mass of the system from the definition of weight, w = mg.\nfind the acceleration of the system from the second law.\nuse kinematics to find the distance moved in 2.00 s, with v0 = 0.\nlearn more\nremarks note that the constant applied force of 20.0 n is assumed to act on the system at all times during its motion. if the force were removed at some instant, the system would continue to move with constant velocity and hence zero acceleration. the rollers have an effect that was neglected here.\nquestion if the weight of the crate were doubled, the acceleration would be multiplied by \nand the displacement attained in 2.0 s would be multiplied by \npractice it\nuse the worked example above to help you solve this problem. the combined weight of the crate and dolly as shown in the figure is 2.90 x 10^2 n. if the man pulls on the rope with a constant force of 19.5 n, what is the acceleration of the system (crate plus dolly), and how far will it move in 2.00 s? assume that the system starts from rest and that there are no friction forces opposing the motion.\nacceleration 0.659 m/s^2\ndisplacement 1.32 m\nenter a number.\nexercise\nhints: getting started | im stuck!\na man pulls a 48.0 kg box horizontally from rest while exerting a constant horizontal force, displacing the box 2.60 meters in 2.00 seconds. find the force the man exerts on the box. (ignore friction.)

solution\nfind the mass of the system from the definition of weight, w = mg.\nfind the acceleration of the system from the second law.\nuse kinematics to find the distance moved in 2.00 s, with v0 = 0.\nlearn more\nremarks note that the constant applied force of 20.0 n is assumed to act on the system at all times during its motion. if the force were removed at some instant, the system would continue to move with constant velocity and hence zero acceleration. the rollers have an effect that was neglected here.\nquestion if the weight of the crate were doubled, the acceleration would be multiplied by \nand the displacement attained in 2.0 s would be multiplied by \npractice it\nuse the worked example above to help you solve this problem. the combined weight of the crate and dolly as shown in the figure is 2.90 x 10^2 n. if the man pulls on the rope with a constant force of 19.5 n, what is the acceleration of the system (crate plus dolly), and how far will it move in 2.00 s? assume that the system starts from rest and that there are no friction forces opposing the motion.\nacceleration 0.659 m/s^2\ndisplacement 1.32 m\nenter a number.\nexercise\nhints: getting started | im stuck!\na man pulls a 48.0 kg box horizontally from rest while exerting a constant horizontal force, displacing the box 2.60 meters in 2.00 seconds. find the force the man exerts on the box. (ignore friction.)

Answer

Explanation:

Step1: Find mass from weight

Given $w = mg$, so $m=\frac{w}{g}$. For the new - case where $w$ is doubled, $m_{new}=\frac{2w}{g}=2m$.

Step2: Find acceleration change

From Newton's second law $a = \frac{F}{m}$. Initially $a_1=\frac{F}{m}$, when mass is doubled ($m_{new} = 2m$), $a_2=\frac{F}{2m}=\frac{1}{2}a_1$. So the acceleration is multiplied by $\frac{1}{2}$.

Step3: Find displacement change

The kinematic equation is $\Delta x=v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0$, $\Delta x=\frac{1}{2}at^{2}$. Initially $\Delta x_1=\frac{1}{2}a_1t^{2}$, when acceleration becomes $a_2=\frac{1}{2}a_1$, $\Delta x_2=\frac{1}{2}a_2t^{2}=\frac{1}{2}(\frac{1}{2}a_1)t^{2}=\frac{1}{2}\Delta x_1$. So the displacement is multiplied by $\frac{1}{2}$.

Answer:

$\frac{1}{2}$, $\frac{1}{2}$