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Question
- a 12 kg sled experiences 30 n of pulling force and 6 n of friction. what is its acceleration?
- a 1,000 kg car accelerates at 3 m/s². what is the net force acting on it?
- a 0.3 kg tennis ball experiences a 1.2 n net force. find its acceleration.
- a 5 kg block slides on a surface with 2 n friction. a 15 n horizontal force is applied. what is its acceleration?
- a 50 kg student jumps and accelerates upward at 1.5 m/s². find the net force acting on the student.
- a 25 kg object is pulled with 100 n of force. if friction is 25 n, what is its acceleration?
- a 10 kg box on a smooth floor is pushed with 40 n. what is the acceleration?
Step1: Determine the net force
According to Newton's second law \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. For problems with friction, the net force \(F_{net}=F_{applied}-F_{friction}\).
Step2: Calculate the acceleration
Rearrange the formula \(a=\frac{F_{net}}{m}\) to find the acceleration.
Problem 17
- Step1: Calculate net force
\(F_{net}=30 - 6=24\space N\)
- Step2: Calculate acceleration
\(a=\frac{24}{12}=2\space m/s^{2}\)
Problem 18
- Step1: Use Newton's second law
\(F = ma\), where \(m = 1000\space kg\) and \(a=3\space m/s^{2}\)
- Step2: Calculate force
\(F=1000\times3 = 3000\space N\)
Problem 19
- Step1: Use Newton's second law
\(a=\frac{F}{m}\), where \(F = 1.2\space N\) and \(m=0.3\space kg\)
- Step2: Calculate acceleration
\(a=\frac{1.2}{0.3}=4\space m/s^{2}\)
Problem 20
- Step1: Calculate net force
\(F_{net}=15 - 2=13\space N\)
- Step2: Calculate acceleration
\(a=\frac{13}{5}=2.6\space m/s^{2}\)
Problem 21
- Step1: Use Newton's second law
\(F = ma\), where \(m = 50\space kg\) and \(a=1.5\space m/s^{2}\)
- Step2: Calculate force
\(F=50\times1.5 = 75\space N\)
Problem 22
- Step1: Calculate net force
\(F_{net}=100 - 25=75\space N\)
- Step2: Calculate acceleration
\(a=\frac{75}{25}=3\space m/s^{2}\)
Problem 23
- Step1: Use Newton's second law
\(a=\frac{F}{m}\), where \(F = 40\space N\) and \(m=10\space kg\)
- Step2: Calculate acceleration
\(a=\frac{40}{10}=4\space m/s^{2}\)
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