Question

the cart in figure has a mass of 2.9 kg and is attached to a 1.7 kg object. calculate the acceleration of the cart given the following assumptions: (a) the force of friction is negligible. (b) the frictional force acting on the wheels of the cart has a magnitude of 0.8 n.

Explanation:

Step1: Analyze the system (cart + object)

The total mass of the system is \( m_{\text{total}} = m_{\text{cart}} + m_{\text{object}} = 2.9\,\text{kg} + 1.7\,\text{kg} = 4.6\,\text{kg} \). The force causing acceleration is the weight of the object, \( F = m_{\text{object}}g \), where \( g = 9.8\,\text{m/s}^2 \).

Part (a): Friction is negligible

Step2: Apply Newton's second law (\( F = ma \))

The net force on the system is the weight of the object (since friction is negligible). So, \( F_{\text{net}} = m_{\text{object}}g = m_{\text{total}}a \). Solving for \( a \):
\( a = \frac{m_{\text{object}}g}{m_{\text{total}}} = \frac{1.7\,\text{kg} \times 9.8\,\text{m/s}^2}{4.6\,\text{kg}} \approx \frac{16.66}{4.6} \approx 3.62\,\text{m/s}^2 \).

Part (b): Friction is \( 0.8\,\text{N} \)

Step3: Adjust net force for friction

The net force is now \( F_{\text{net}} = m_{\text{object}}g - f \), where \( f = 0.8\,\text{N} \).
\( F_{\text{net}} = 1.7\,\text{kg} \times 9.8\,\text{m/s}^2 - 0.8\,\text{N} = 16.66\,\text{N} - 0.8\,\text{N} = 15.86\,\text{N} \).
Using \( F_{\text{net}} = m_{\text{total}}a \), solve for \( a \):
\( a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{15.86\,\text{N}}{4.6\,\text{kg}} \approx 3.45\,\text{m/s}^2 \).

Answer:

(a) Acceleration (friction negligible): \(\approx 3.6\,\text{m/s}^2\) (or more precisely \( \approx 3.62\,\text{m/s}^2 \))
(b) Acceleration (with friction): \(\approx 3.5\,\text{m/s}^2\) (or more precisely \( \approx 3.45\,\text{m/s}^2 \))