Question

homework 3 begin date: 8/17/2025 11:59:00 pm due date: 9/9/2025 11:59:00 pm end date: 9/9/2025 11:59:00 pm problem 7: (6% of assignment value) a block with a mass of m = 44 kg rests on a frictionless surface and is subject to two forces acting on it. the first force is directed in the negative x-direction with a magnitude of f₁ = 9.5 n. the second has a magnitude of f₂ = 23 n and acts on the body at an angle θ = 12° measured from horizontal, as shown. part (a) ✔ write an expression for the component of net force, f_net,x, in the x-direction, in terms of the variables given in the problem statement. f_net,x = -f₁ + f₂ cos(θ) ✔ correct! part (b) find the blocks acceleration in the x-direction, a_x, in meters per second squared. a_x = |

Explanation:

Step1: Calculate net force in x-direction

Using the expression from part (a): \( F_{\text{net},x} = -F_1 + F_2 \cos(\theta) \)
Substitute \( F_1 = 9.5 \, \text{N} \), \( F_2 = 23 \, \text{N} \), \( \theta = 12^\circ \)
\( F_{\text{net},x} = -9.5 + 23 \cos(12^\circ) \)
Calculate \( \cos(12^\circ) \approx 0.9781 \)
\( F_{\text{net},x} = -9.5 + 23 \times 0.9781 \)
\( F_{\text{net},x} = -9.5 + 22.4963 \)
\( F_{\text{net},x} \approx 12.9963 \, \text{N} \)

Step2: Use Newton's second law to find acceleration

Newton's second law: \( F_{\text{net},x} = m a_x \)
So, \( a_x = \frac{F_{\text{net},x}}{m} \)
Substitute \( F_{\text{net},x} \approx 12.9963 \, \text{N} \), \( m = 44 \, \text{kg} \)
\( a_x = \frac{12.9963}{44} \approx 0.2954 \, \text{m/s}^2 \)

Answer:

\( \approx 0.30 \, \text{(or more precise } 0.295 \text{)} \)