Question

a block of mass 25 kg slides up a frictionless plane inclined at 11° with the horizontal under the influence of a force of 140 n acting parallel to the incline and up the incline. the acceleration of gravity is 9.8 m/s². what is the acceleration of the block? answer in units of m/s². 1. 4.68996 2. 2.79548 3. 3.14641 4. 5.03933 5. 5.38039 6. 3.73007 7. 9.33964 8. 2.44959 9. 5.53748 10. 7.24371

Explanation:

Step1: Analyze forces along the incline

According to Newton's second law \(F_{net}=ma\). The net force along the incline is \(F - mg\sin\theta\), where \(F = 140N\), \(m = 25kg\), \(g=9.8m/s^{2}\), and \(\theta = 11^{\circ}\).
So, \(F_{net}=140 - 25\times9.8\times\sin(11^{\circ})\)
First, calculate \(\sin(11^{\circ})\approx0.1908\)
Then \(25\times9.8\times0.1908 = 25\times1.86984=46.746\)
\(F_{net}=140 - 46.746 = 93.254\)

Step2: Calculate acceleration

Since \(F_{net}=ma\), then \(a=\frac{F_{net}}{m}\)
Substitute \(F_{net}=93.254N\) and \(m = 25kg\)
\(a=\frac{93.254}{25}=3.73016\approx3.73007\)

Answer:

1. 3.73007