Question

for a moving object, the force acting on the object varies directly with the objects acceleration. when a force of 16 n acts on a certain object, the acceleration of the object is 4 m/s². if the acceleration of the object becomes 9 m/s², what is the force?

Explanation:

Step1: Find the mass of the object

According to Newton's second law \(F = ma\). When \(F = 16\space N\) and \(a\) is the initial acceleration. Let the mass of the object be \(m\). Then \(m=\frac{F}{a}\). Since \(F = 16\space N\) (we assume the initial force - acceleration relation is \(F = ma\) and we can find \(m\) first).

Step2: Calculate the new force

When the acceleration \(a'=9\space m/s^{2}\), and mass \(m\) is constant. From \(F = ma\), when \(F_1 = 16\space N\) and \(a_1\) (initial acceleration, assume \(a_1\) is the acceleration corresponding to \(F_1\)), \(m=\frac{F_1}{a_1}\). But if we assume the first case \(F_1 = 16\space N\) and \(a_1\) (say \(a_1\) is the acceleration when \(F = 16\space N\)), and since \(F=ma\), \(m=\frac{16}{a_1}\). When \(a' = 9\space m/s^{2}\), \(F'=m\times a'\). Since \(F\propto a\) (because \(m=\frac{F}{a}\) is constant), we can also use the proportion \(\frac{F_1}{a_1}=\frac{F_2}{a_2}\). Assuming \(a_1\) is the acceleration when \(F_1 = 16\space N\) and \(a_2 = 9\space m/s^{2}\). But if we assume the first - case \(F_1 = 16\space N\) and \(a_1\) (say \(a_1\) is the acceleration when \(F = 16\space N\)), and since \(F = ma\), \(m=\frac{16}{a_1}\). When \(a'=9\space m/s^{2}\), \(F'=m\times a'\). Another way: Since \(F = ma\), and \(m\) is constant. Let \(F_1 = 16\space N\), \(a_1\) (initial \(a\)), \(F_2\) (unknown), \(a_2=9\space m/s^{2}\). From \(F = ma\), \(m=\frac{F_1}{a_1}=\frac{F_2}{a_2}\). If we assume the first - case \(F_1 = 16\space N\) and \(a_1\) (say \(a_1\) is the acceleration when \(F = 16\space N\)), and since \(F\propto a\) (because \(m\) is constant). Let's first find \(m\) from \(F = ma\). When \(F = 16\space N\), assume \(a_1\) (initial \(a\)): \(m=\frac{16}{a_1}\). When \(a = 9\space m/s^{2}\), \(F=ma\). But if we use the direct proportion (because \(m=\frac{F}{a}\) is constant, so \(F\) and \(a\) are directly proportional). Let \(F_1 = 16\space N\), \(a_1\) (initial \(a\)), \(F_2\) (unknown), \(a_2 = 9\space m/s^{2}\). From \(\frac{F_1}{a_1}=\frac{F_2}{a_2}\). If we assume the first - case \(F_1 = 16\space N\) and \(a_1\) (say \(a_1\) is the acceleration when \(F = 16\space N\)), and since \(F = ma\), \(m\) is constant. Let's use \(F=ma\). First, find \(m\) from \(F = 16\space N\) (assuming \(a_1 = 4\space m/s^{2}\) from the incomplete problem description, maybe a mis - cut. If \(a_1 = 4\space m/s^{2}\), then \(m=\frac{F_1}{a_1}=\frac{16}{4}=4\space kg\)). Then when \(a = 9\space m/s^{2}\), \(F=ma\). Substitute \(m = 4\space kg\) and \(a = 9\space m/s^{2}\) into \(F=ma\).

Answer:

36