a = F(net)/M = (80 - 5)/M
= 75/3 = 25 m/s^2
A 3.0 kg block being pulled across a table by a horizontal force of 80 N also experiences a frictional force of 5 N. What is the acceleration of the block?
2 answers
f1= 80N ; f2=5N
F=f1 - f2 = 80N - 5N= 75N
F=M*a
a= F/M =75N/3kg =25 m/s^2
a=25 m/s^2
F=f1 - f2 = 80N - 5N= 75N
F=M*a
a= F/M =75N/3kg =25 m/s^2
a=25 m/s^2
6 answers