What is the Unknown Force Acting on a 2.8 kg Object in Upward Vertical Motion?

  • Thread starter Thread starter tdusffx
  • Start date Start date
  • Tags Tags
    Force Vertical
AI Thread Summary
A 2.8 kg object in upward vertical motion experiences an unknown force in addition to its weight. The object, starting from rest, has a displacement of (4.2i - 3.3j) m in 1.2 seconds. To find the other force, the displacement is split into i and j components, and the total force is calculated using the object's acceleration. The weight force is then subtracted from the total force to determine the unknown force. The discussion highlights the importance of correctly handling signs in force calculations.
tdusffx
Messages
58
Reaction score
0
can anyone help me start this problem?

Besides its weight, a 2.8 kg object is subjected to one other constant force. The object starts from rest and in 1.2s experieces a displacement of (4.2i - 3.3j) m, where the direction of j is the upward vertical direction. determine the other force
 
Physics news on Phys.org
Start by splitting the displacement into i and j components. Using the displacements calculate the total force on the object. Then subtract the weight force from the total force to get the 'other force'.
 
kk, I got...

deltaX = Vixt + 1/2at^2

4.2 = 0 1/2a(1.2)^2

a = 5.8i

deltaY = Viyt + 1/2at^2

-3.3 = 0 + 1/2a(1.2)^2

Y = -5.4j

F = 2.8kg(5.8i + 5.4j) ?
 
Seems roughly ok. I don't have to check your arithmetic, right? But aren't you losing signs again in the total force? Shouldn't the y component of the force be DOWN?
 
Last edited:
yup yup! you're right again, dick. Signs are always messing me up. Thanks again.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Equilibrium of Forces and Torques on Sawhorses Supporting a Person on a Board
Replies
6
Views
2K
Forces on an object in vertical circular motion
Replies
8
Views
2K
Vertical circle in a pendulum ride -- tension force acting on the gondola
Replies
7
Views
2K
Distribution of Energy when work is done on a system of 2 masses connected by a spring
Replies
17
Views
2K
Bullet shooting vertically into a block
Replies
9
Views
7K
Forces - A block slides upward on a rough, vertical wall
Replies
16
Views
7K
What is the induced magnetic force on this closed current loop?
Replies
12
Views
2K
Force Direction: Understanding Upward Motion on Lowering Objects
Replies
2
Views
1K
How Do Forces Affect Particle Motion in Different Vertical Circle Scenarios?
Replies
25
Views
2K
Object Acceleration with Vertical Forces: Rising or Descending?
Replies
9
Views
2K

Hot Threads

Recent Insights

Back
Top