A block has weight W and mass m, is pulled by tension T at angle θ with the horizon, and experiences a force of friction, Fk, opposing its motion across a rough surface. If the box moves a distance d, what is the acceleration of the box?
a) Fkd/W
b) (T-W-Fk)g/W
c) (Tcosθ – Fk)g/W
d) (Tsinθ – Fk>)/W
Explanation
The acceleration of the box is equal to the sum of forces in the horizontal direction divided by the mass of the box. The sum of horizontal forces is Tcosθ – Fk, and the mass in terms of weight is W/g. The distance is unnecessary for the solution.
ΣFx=max
Tcosθ – Fk=max
Tcosθ – Fk=W/gax
ax=(Tcosθ – Fk)g/W
Want more MCAT practice?
We’ve got options for every schedule and learning style!
From the best online MCAT course created by top instructors with 524+ MCAT scores to the most representative full-length practice exams and private tutoring, we can custom tailor your MCAT prep to your goals!
Not sure which option is right for you? Schedule a free MCAT consultation with an MCAT expert using the form below. No obligation, just expert advice.
Related Posts
Search the Blog
Free Consultation
Interested in our Online MCAT Course, One-on-One MCAT Tutoring or Med admissions packages? Set up a free consultation with one of our experienced Senior Student Advisors.
Schedule NowPopular Posts
-
-
MCAT Blog What's on the MCAT? -
-
-
MCAT Blog How to Review MCAT Full Lengths
Free MCAT Practice Account
Need great MCAT practice?Get the most representative MCAT practice possible when you sign up for our free MCAT Account, which includes a half-length diagnostic exam and one of our full-length MCAT practice exams.
Learn More