for a 3.0 s, alex exerts a net force of 10.0 n on a 6.7 kg shopping cart that was initially at rest. use the…

for a 3.0 s, alex exerts a net force of 10.0 n on a 6.7 kg shopping cart that was initially at rest. use the above information to answer questions 24 - 26.\n32. the acceleration of the cart is\n(a) 9.8 m/s²\n#(b) 1.49 m/s²\n(c) 5.6 m/s²\n(d) none of the above\n33. the distance traveled by the cart is\n(a) 8.65 m\n(b) 4.94 m\n(c) 5.98 m\n(d) 6.75 m\n(e) none of the above\n34. the final velocity of the cart\n(a) 4.47 m/s

for a 3.0 s, alex exerts a net force of 10.0 n on a 6.7 kg shopping cart that was initially at rest. use the above information to answer questions 24 - 26.\n32. the acceleration of the cart is\n(a) 9.8 m/s²\n#(b) 1.49 m/s²\n(c) 5.6 m/s²\n(d) none of the above\n33. the distance traveled by the cart is\n(a) 8.65 m\n(b) 4.94 m\n(c) 5.98 m\n(d) 6.75 m\n(e) none of the above\n34. the final velocity of the cart\n(a) 4.47 m/s

Answer

Explanation:

Step1: Find acceleration using Newton's second - law

According to $F = ma$, where $F$ is force, $m$ is mass and $a$ is acceleration. So $a=\frac{F}{m}$. Given $F = 10.0\ N$ and $m = 6.7\ kg$, then $a=\frac{10.0}{6.7}\approx1.49\ m/s^{2}$.

Step2: Find distance using kinematic equation

The kinematic equation for distance with initial velocity $u = 0\ m/s$, acceleration $a$ and time $t$ is $s=ut+\frac{1}{2}at^{2}$. Since $u = 0\ m/s$, $a = 1.49\ m/s^{2}$ and $t = 3.0\ s$, then $s=\frac{1}{2}\times1.49\times3.0^{2}=\frac{1}{2}\times1.49\times9 = 6.705\approx6.75\ m$.

Step3: Find final velocity using kinematic equation

The kinematic equation $v=u + at$. Given $u = 0\ m/s$, $a = 1.49\ m/s^{2}$ and $t = 3.0\ s$, then $v=0+1.49\times3.0 = 4.47\ m/s$.

Answer:

  1. B. $1.49\ m/s^{2}$
  2. D. $6.75\ m$
  3. A. $4.47\ m/s$