Question

question 47
two forces act on the 7.00 kg block shown in the figure. the magnitudes of the forces are ( f_1 = 45.0 , \text{n} ) and ( f_2 = 25.0 , \text{n} ). what is the magnitude of the horizontal acceleration of the block?

( 5.77 , \text{m/s}^2 )
( 2.20 , \text{m/s}^2 )
( 1.37 , \text{m/s}^2 )
( 10.0 , \text{m/s}^2 )
( 2.86 , \text{m/s}^2 )

Explanation:

Step1: Calculate the horizontal component of \( F_1 \)

The horizontal component of \( F_1 \) is \( F_{1x}=F_1\cos\theta \), where \( F_1 = 45.0\ N \) and \( \theta = 70.0^{\circ} \).
\( F_{1x}=45.0\times\cos(70.0^{\circ})\)
Using \( \cos(70.0^{\circ})\approx0.342 \), we get \( F_{1x}=45.0\times0.342 = 15.39\ N \)

Step2: Find the net horizontal force

The net horizontal force \( F_{net}=F_2 - F_{1x} \), with \( F_2 = 25.0\ N \)
\( F_{net}=25.0 - 15.39=9.61\ N \)

Step3: Use Newton's second law \( F = ma \) to find acceleration

Given \( m = 7.00\ kg \), from \( a=\frac{F_{net}}{m} \)
\( a=\frac{9.61}{7.00}\approx1.37\ m/s^{2} \)

Answer:

\(1.37\ m/s^{2}\) (corresponding to the third option)