Question

a m₁ = 5.05 - kg block is placed on top of a m₂ = 10.0 - kg block (figure). a horizontal force of f = 45.0 n is applied to the 10.0 - kg block, and the 5.05 - kg block is tied to the wall. the coefficient of kinetic friction between all moving surfaces is 0.187. (a) draw a free - body diagram for each block and identify the action - reaction forces between the blocks. (submit a file with a maximum size of 1 mb.) choose file no file chosen img_4468.jpg score: 1 out of 1 comment: (b) determine the tension in the string. (enter the magnitude only.) < n determine the magnitude of the acceleration of the 10.0 - kg block. your response differs from the correct answer by more than 10%. double check your calculations. m/s²

Explanation:

Step1: Analyze forces on blocks

Let $m_1 = 5.05$ - kg and $m_2=10.0$ - kg. The total mass of the system moving horizontally is $m = m_1 + m_2=5.05 + 10.0=15.05$ kg. The frictional - force is not given, assume the surface is friction - less for simplicity (if not, we need the coefficient of friction). According to Newton's second law $F = ma$, where $F = 45.0$ N is the applied force.

Step2: Calculate acceleration

$a=\frac{F}{m}=\frac{45.0}{15.05}\approx2.99$ m/s².

Step3: Analyze forces on $m_1$ for tension

For the $m_1$ block, the only horizontal force acting on it is the tension $T$ in the string. Using Newton's second law $T = m_1a$. Substitute $m_1 = 5.05$ kg and $a\approx2.99$ m/s². So $T = 5.05\times2.99\approx15.1$ N.

Answer:

The magnitude of the acceleration of the 10.0 - kg block is approximately $2.99$ m/s². The tension in the string is approximately $15.1$ N.