Question

a block is acted on by two forces as shown in the diagram below.oscophy2018 4.7 v
if the magnitudes of the forces are ( f _ { 1 } = 51.0 mathrm { n } ) and ( f _ { 2 } = 20.5 mathrm { n } ), what are the magnitude (in ( mathrm { m } / mathrm { s } ^ { 2 } )) and direction of the acceleration of the block? let ( m = 8.00 mathrm { kg } ) and ( \theta = 37.5 ^ { circ } ).
magnitude ( mathrm { m } / mathrm { s } ^ { 2 } )
direction

Explanation:

Step1: Calculate the net force in the x - direction

The x - component of \(F_{2}\) is \(F_{2x}=-F_{2}\cos\theta\) (negative because it is in the opposite direction of \(F_{1}\)). The net force \(F_{netx}=F_{1}+F_{2x}=F_{1}-F_{2}\cos\theta\).
Substitute \(F_{1} = 51.0\space N\), \(F_{2}=20.5\space N\), and \(\theta = 37.5^{\circ}\)
\(F_{netx}=51.0-20.5\cos(37.5^{\circ})\)
\(\cos(37.5^{\circ})\approx0.7939\)
\(F_{netx}=51.0 - 20.5\times0.7939=51.0-16.27495 = 34.72505\space N\)

Step2: Use Newton's second law \(F = ma\) to find the acceleration

We know that \(a=\frac{F_{netx}}{m}\), where \(m = 8.00\space kg\)
\(a=\frac{34.72505}{8.00}=4.34063125\space m/s^{2}\approx4.34\space m/s^{2}\)
The direction is the same as the direction of \(F_{1}\) (to the right) since \(F_{netx}>0\)

Answer:

The magnitude of the acceleration is \(4.34\space m/s^{2}\) and the direction is the same as the direction of \(\vec{F}_{1}\) (to the right).