Finding the weight, work, and normal force of a block

AI Thread Summary
To find the weight of a block, multiply its mass by the acceleration due to gravity (g = 10 m/s²). The work done by the weight can be calculated using the formula W = Fd, considering the direction of the force. The normal force acting on a block can be determined by analyzing the forces in the vertical direction, factoring in weight and friction. Understanding Newton's second law (F=ma) is crucial for solving these problems. Clarification on vector products and integrals may enhance comprehension of the concepts involved.
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Homework Statement


4 questions:
1. A block of mass M=2.6 kg is dragged over a horizontal surface by a force F=22.0 N. The block is displaced a distance d=10.0 m, the friction force Ff is 3.1 N. What is the weight of the block in Newtons? Use 10 m/s2 for g.

2. A block of mass M=1.0 kg is dragged over a horizontal surface by a force F=20.0N. The block is displaced a distance d=10.0 m, the friction force Ff is 5.0 N. What is the work done by the weight Fg?

3. A block of mass M=2.6 kg is dragged over a horizontal surface by a force F=22.0 N. The block is displaced a distance d=10.0 m, the friction force Ff is 3.1 N. What is the normal force acting on the block?

4. A block of mass M=3.3 kg is dragged over a horizontal surface by a force F=29.0 N. The block is displaced a distance d=17.0 m, the friction force Ff is 4.8 N. What is the work done by the normal force FN?


Homework Equations



W=fxd


The Attempt at a Solution



I have tried to watch my instructors videos and none of this section makes any sense to me! I am in need of someone to explain how to figure all of these out please !
Thank you so much
 
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Hi,
The fundamental thing to remember here, is Newton's second law, or F=ma(mass times acceleration).
Weight is a special case of a force, where the acceleration is a constant g, and it is always downward, depending on your choice of axes.
You wrote down correctly, that \large W = Fd, but remember that it's a vector product, a dot product, and preferably an integral, W = \displaystyle \int \vec{F}\cdot d\vec{r}.
Does this make things clearer?
Daniel
 
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