At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
The frictional force is 11 N and the block’s acceleration is 0.14 m/s².
What is the coefficient of friction?
It is defined as the numerical value that indicates the amount of friction present between the surfaces of two bodies. The lower the coefficient of friction, the lower the friction between the surfaces, and the higher coefficient of friction the higher the friction force between them.
For part (a):
Taking x component:
F(x) = F cosθ
Taking y component:
F(y) = -F sinθ
Applying Newton’s second law to the y-axis:
F(N) = 15 sin40 + (3.5)(9.8)
F(N) = 44 N
coefficient U = 0.25
f(k) = 11 N
For part (b):
Applying Newton’s second law to the x-axis:
a = ((15) cos40 - 11)/3.5
a = 0.14 m/s²
Thus, the frictional force is 11 N and the block’s acceleration is 0.14 m/s².
Learn more about the coefficient of friction here:
brainly.com/question/13828735
#SPJ4
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.